IIT_Class_XI_Phy_Rotation Motion own edition.pdf

   S T U D Y M A T E R I A L          ROTATIONAL MOTION    IIT-JEE             

PHYSICS

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C O N T E N T S

CONTENTS

ROTATIONAL MOTION 1.

Theory

2.

Solved Problems (Subjective, Objective, Multiple Choice, Passage Based, Matching , Assertion-Reason)

3.

Assignments Section - I : Subjective Questions (Level I, Level - II and Level - III) Section - II : Single Choice Questions Section - III : Multiple Choice Questions Section - IV •

Passage Based Questions

•

Matching Type Questions

•

Assertion-Reason Type Questions

Section - V : Problems Asked in IIT-JEE 4.

Answers

ROTATIONAL MOTION IIT-JEE-Syllabus Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres Equilibrium of rigid bodies; Collision of point masses with rigid bodies

RIGID BODIES AND ROTATION A body with a definite and unchanging shape and size is a rigid body. In practice such a body does not exist in nature. A real body undergoes changes in shape and size under the action of external forces. If however, these forces are small enough and we are dealing with solid bodies, changes in shape or size are small enough. Under these circumstances, we may regard the body as a rigid body. The motion of a rigid body in three dimensions is, in general, very complicated. The analysis of this motion can be done in terms of two simple types of motion: translation and rotation, a classification which is based on the spatial properties of motion. We define these two types of motion: (a)

Translation: A rigid body is said to undergo pure translation if it moves such that all its particles undergo the same displacement during any interval of time. The body remains parallel to itself throughout its motion i.e. a straight line joining any two particles remains parallel to itself throughout its motion (see figure). Note that the motion need not be confined to a plane, i.e. particles of the body may move out of the plane and the motion is still translation so long as it remains parallel to itself.

(b)

A2 B2

A B

A1 B1

Rotation: A rigid body is said to undergo pure rotation if it moves such that its constituent particles move along circular arcs, the centres of which lie on a straight line perpendicular to the plane of the arcs. This straight line is known as the axis of rotation, and the rigid body is said to undergo pure rotation about this axis.

Figure : “Translation preserves orientation” AB = A2B2 (rigid body) AB || A2B2 (translation) ∴ ABB2A2 is a parallelogram

JJJJG JJJG ⇒ AA 2 = BB 2 .

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In the following figures we give two examples: one of translation and one of rotation. One should try to understand how they fit in with the definitions. A2

A1 B1

B2

A2

A

B2 B

B1

A0

A1

A3 A0

B3

B0 B4 A4

AB || A1B1 || A2B2 Figure: Orientation is preserved and therefore the motion is translation. Note that the circular motion of A, B take place with centres at A0, B0 respectively: but their centres don’t lie on a straight line perpendicular to the plane of motion.

Figure: Orientation is not preserved. Note that particles A and B move along concentric circles in the same plane. The motion is rotation about an axis perpendicular to plane of the paper and passing through

A0 .

The axis of rotation may be fixed in space (stationary axis) or moving (changing in position or direction). In this chapter, we are going to discuss the rotation of a rigid body about a fixed axis, and also about an axis which may be moving but without any change in its orientation (pure rolling). The following figures show both types of rotation:

Figure:

Rotation about a stationary axis: A fixed pulley with a falling block attached to it by means of a string.

Figure:

Rotation about a translating axis: A cylinder rolling on an inclined plane.

The physics of rotational motion may be studied using the following kinematic variables: (i) orientation, measured by an angular variable, θ (ii) the angular velocity, ω , and, (iii) the angular acceleration, α .

ROTATIONAL KINEMATICS Let us consider the rotation of a rigid body about a fixed axis, i.e. w.r.t. an axis that is fixed in some arbitrary frame of reference (not necessarily inertial). y

We take a convenient point O on this axis, choosing this as our origin and the axis of rotation itself as the z-axis, we take another pair of axes x and y, forming a triad of mutually perpendicular lines intersecting at O. Conventionally, the three axes are chosen so as to form a right handed system of coordinates. We take a particle, P, of the rigid body in the x-y plane. Since the body undergoes rotation about the z-axis, the particle P rotates in a circle centred at O with the radius OP.

P

O

θ

x

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(i)

Angular Position and Displacement: The angle made by OP with the x-axis is denoted by θ , and this specifies the orientation of the rigid body. It is, in general, a function of time. If it is chosen to be zero at the initial instant, it is referred to as the angular displacement. Angular displacement is a very important quantity. It is the same for all particles of a rigid body during rotation. Thus the rotational motion of all the particles of a rigid body may be described using a single angular variable θ. It is measured in the anticlockwise sense with respect to the axis of rotation, and its unit is radian. Angular displacement (or position) is not, in general, a vector quantity as it does not obey the rules of vector addition. Infinitesimal angular displacements are, however, vectors.

(ii)

Angular Velocity : The average angular velocity, ωav , is defined as ωav =

θ2 − θ1 t 2 −t 1

where, θ2 and θ1 represent the orientations of the body at times t 2 and t 1 (t 2 > t 1 ) respectively. The instantaneous angular velocity, ω is defined by, ∆θ ω = lim ∆t → 0 ∆t θ (t + t ) − θ ( t ) dθ = lim = . ∆t → 0 ∆t dt →

The angular velocity vector ω may be defined by : → ω = d θ nˆ , where nˆ is the unit vector directed along the axis of rotation (in this case : the z-axis). dt The angle θ is measured in the anticlockwise sense, in a plane perpendicular to the vector nˆ , using the right-hand-screw rule. (iii)

Angular Acceleration : The average angular acceleration is defined by G G ω (t 2 ) − ω (t 1 ) G αav = t 2 − t1 G where ω (t) represents the instantaneous angular velocity at time t. The instantaneous angular acceleration is defined using calculus; by the expression: G G dω . α= dt All the quantities referred to above are defined with respect to an axis fixed in direction.

COMPARISON BETWEEN TRANSLATIONAL AND ROTATIONAL KINEMATICS The kinematic equations in case of linear and rotational motion are compared in the following table: L

Linear motion of particle G G dr v = dt G G dν a= dt

Rotational Motion (fixed axis)

G dθ ω= nˆ dt G G dω α= nˆ dt Motion under uniform acceleration (a = constant, α = constant) v = u + at ω = ωo + αt 1 2 1 x = ut + at θ = ω0t + αt 2 2 2 v 2 = u 2 + 2a x ω2 = ω2 + 2 αθ . 0

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Illustration 1.

A wheel rotates with a constant acceleration of 4.0 rad/s2 starting with an initial angular velocity of 2.0 rad/s. If the wheel starts from rest, how many revolutions will it make in the first 5 seconds?

Solution :

The angular displacement in the first 5 seconds is given by: 1 2 1 αt = 2.0 × 5 + (4.0 rad/s2) (5s)2 = 60 rad. 2 2 The wheel turns by 2 π radians in each revolution, the number of revolutions in 5 s is 60 n= = 9.6 (approx). 2π θ = ω0t +

Illustration 2.

Starting from rest, a fan takes ten seconds to attain the maximum speed of 600 rpm (revolutions per minute). Assuming constant acceleration, find the time taken by the fan in attaining half the maximum speed.

Solution :

Let the angular acceleration be α . According to the question,

Illustration 3.

Solution :

600 rev/min = 0 + α 10. … (i) Let t be the time taken in attaining the speed of 300 rev/min which is half the maximum. Then, 300 rev/min = 0 + αt. … (ii) Dividing (i) by (ii), we get, 10 or, t = 5 s. 2= t A body rotates with an angular acceleration which is linearly proportional to its angular displacement (θ): α = λθ, the proportionality constant λ being positive. Find the angular velocity of the body as a function of angular displacement and the angular displacement as a function of time. It is given displacement as a function of time. It is given that ω = ω0 and θ = θ0 at t = 0; where ω0 = λθ0 . It is given that: α = λθ d ω d ωd θ dω Now, α = = =ω . dt d θ dt dθ dω ∴ ω = λθ . dθ ω2 θ2 = λ +C 1 . Integrating, we get, 2 2 For t = 0, ω20 θ2 = λ 0 +C 1 . 2 2 Using the given condition, C1 = 0 ω2 = λθ2 ∴ or, ω = λθ . dθ To find θ = θ (t), we write ω = dt dθ = λθ dt θ

t

or,

dθ ∫θ θ = λ ∫0 dt 0

or,

θ = θ0e

λt

or, ln

θ = λt θ0

.

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Illustration 4.

In torsional oscillations, it is observed that the angular acceleration α, is proportional to θ, the angular displacement but directed opposite: α = – λθ, λ = a positive constant. Integrate the above expression, retaining the arbitrary constants and obtain θ as a function of time t.

Solution :

This is similar to the previous illustration in that the steps are identical. dω ω= = −λθ . dθ Separating variables and integrating, ω2 λ = − θ2 + C 1, where C1 is the constant of integration. 2 2 λθ20 C1 is written in the form: . 2 The expression for ω becomes ω2 = λ( θ02 − θ2 ) ω = ± λ θ20 − θ2 . Taking the positive root (the case with the negative root is being left as an exercise for the student), we write, dθ = λ θ20 − θ2 dt dθ or, ∫ θ2 − θ2 = λ ∫ dt 0 or,

Illustration 5.

Solution :

θ = λt + φ0 , where φ0 is the constant of integration. θ0

or,

sin −1

∴

θ = θ0 sin( λt + φ0 ) .

A body is rotating with an angular retardation proportional to the square of its angular speed: α = −b ω2 , b > 0. (a) (b)

Find its angular speed when it has turned through one complete revolution. The ratio of the times taken to complete the 1st and the 2nd revolution is t1 1 = . Find the value of b. t2 2

(a)

Angular acceleration α =

d ω d ωd θ dω = =ω . dt dt dt dθ

dω = −b ω2 dθ dω or, = −bd θ . ω ∴ω

Integrating, [ ln ω]ω = − [b θ]0 ω

θ

0

∴ ln

ω = −b θ ω0

∴ ω = ω0e − b θ . For one rotation θ = 2π , hence from equation (i)

… (i)

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(b)

From equation (i), ∴ dt =

dθ = ω0e −b θ dt

1 bθ dθ = e dθ ω0e − b θ ω0

2π

∴ t 1 = ∫ e b θd θ = 0

and t 2 =

1 ω0

4π

∫e

1 2 πb (e − 1) ω0

bθ

dθ=

2π

1 4 πb − e 2 πb ) (e ω0

2 πb

t1 −1 e 1 = 2 πb 2 πb = 2 πb . t 2 e (e − 1) e t1 1 But = (given) t2 2 ∴

1 1 ∴ = 2 πb 2 e or, e 2 πb = 2 2 πb = ln 2 or, b =

1 ln 2 . 2π

RELATIONSHIP BETWEEN LINEAR AND ANGULAR VARIABLES Suppose that particle P of a rigid body undergoing pure rotation about an axis nˆ , moves from its location P at time t to P′ at the time t + dt . The distance moved by P in time dt is, ds = arc PP'′

n′ P′

dθ A P = AP .d θ where, AP is the distance of P from the axis of rotation; the centre of the circle of rotation of P being A. ds = rp dθ β ds dθ vP = = rp = rp ω . O dt dt If the origin of coordinates be chosen at some point O on the axis of rotation, then v p can be written after using the expression, AP = OP sin β vp = OP sin β.ω G v p is directed into the plane and while ω is along the axis, so we can write, G JJG G v P = ω ×r P G where r p is the position vector of P relative to the origin O, on the axis. This expression gives the velocity of P with respect to the axis of rotation. v P = O P sin β.ω If the direction of the axis of rotation is fixed, we can differentiate with respect to time t, to get the linear acceleration of P, JJG G G JJG G JJG G d ω G   JJG d r P  a P =  × r P  +  ω ×  = α × r p + ω ×v P , dt   dt   JJG G where α represents the angular acceleration v P and has been defined in the two previous sections.

(

) (

)

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The relations v = rω and a = rα are very useful and their meanings should be clearly understood. Different particles of the rigid body, have different radii for their circles of rotation, but the same values for ω & α. . Thus, the tangential speed and the tangential acceleration of different particles are different. For r = 0 i.e., for the particles on the axis, ν = rω = 0 and a = rα = 0, since the particles on the axis do not move at all relative to the axis itself.

ROTATIONAL DYNAMICS Kinetic Energy and moment of Inertia: Suppose that a rigid body is undergoing pure rotation about a fixed axis in an inertial frame of reference, with an angular velocity ω. The particles of the rigid body are numbered from 1, 2, 3, ..., n. Let the mass of the ith particle be given by m i and its distance from the axis of rotation be given by ri, then the speed of the particle is : v i = ωri and its kinetic energy is : 1 1 2 m iv 2i = m i ( ωr i ) . 2 2 The kinetic energy of the entire body is : 1 K.E = ∑ m i (ωri ) i 2

ω

K.E. =

ri mi

P

1 2  2  ∑ m i ri  ω 2 i  1 2 = Iω . 2 I = ∑ m i ri 2 , is defined as the moment of inertia of the body about the given axis. =

where

i

For a discrete distribution, the moment of inertia I, is given by: I = ∑ m i ri 2 i

For continuous mass distribution, the moment of inertia, I, is given by I = ∫ r 2d m where r is the distance of the mass-element dm from the axis of rotation. The integration is carried out over the entire mass distribution. Illustration 6.

A nitrogen molecule which consists of two nitrogen atoms separated by a distance 1.3 × 10 −10 m , has an average rotational kinetic energy of about 4 × 10 −21 J . Find the moment of inertia of the nitrogen molecule about its centre of mass and its angular frequency of rotation. Given mass of a nitrogen atom = 14 u 1u = 1.67 ×10 −27 kg .

(

Solution :

)

m = 14 u = 14 × 1.67 × 10 −27 kg = 2.338 × 10 −26 kg . The interatomic separation, a = 1.3 × 10 −10 m and the rotational kinetic energy, KE = 4 × 10 −21 J . The two atoms are identical, therefore the centre of mass is at a a from either atom. Therefore, the moment of distance 2 inertia of a nitrogen molecule about its centre of mass is

1.3 ×10 −10 m

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2

Since,

Illustration 7. Solution :

2 1 ma 2 a  I = 2m   = = × 2.338 ×10 −26 × 1.3 ×10 −10 = 1.98 ×10 -46 kg m 2 . 2 2 2 1 2 KE = I ω , 2 2KE ω= = 6.36 ×1012 rad s-1 (substituting the values). I

(

)

Three point - masses 2 kg, 3 kg and 7 kg are located at the vertices of an equilateral triangle of side 6 m. What is the moment of inertia of the system about an axis along an altitude of the triangle passing through the 2 kg mass? Moment of inertia about altitude AD is I = m 1 × ( distance of m 1 from AD ) + m 2 2

× ( distance of m 2 from A D )

+m 3 × ( distance of m 3 from AD ) = m 1 × 0 + m 2 × ( BD ) + m 3 × (CD ) 2

2

A

2

m1 = 2kg

2

6m

2

2

a  a  = 0 + m2 ×  + m3   2   2 2 a 62 = ( m 2 + m 3 ) = ( 3 + 7 ) = 90 kg m 2 . 4 4

B m2 = 3kg

D

C m3 = 7kg

Radius of Gyration: The radius of gyration of a body about a certain axis is given by k, which is related to the moment of inertia I about the same axis by the relation, I = mk 2 . Illustration 8. Find the radius of gyration of a uniform rod of length A and mass m about an axis passing through the centre of mass of the rod and making an angle θ with the rod. Solution :

The moment of inertia of the rod about the given axis can be easily found by integration : I=

A 2

2 m  dx  ( x sin θ) A  A

∫ 

−

=

1 m A2 sin 2 θ = m k 2 12

2

A sin θ k = . 2 3

Parallel axis Theorem : The moment of inertia of a rigid body about any given axis equals the moment of inertia about a parallel axis passing through its centre of mass plus the moment of inertia of a particle of equal mass about the given axis : I axis = I CM + Md 2 . Parallel axis

Axis of rotation

Mass = M CM d

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Illustration 9.

Four solid spheres, each of mass m and radius r are located with their centres on four corners of a square of side a. What is the moment of inertia of the system about an axis (a) along any diagonal and (b) along any side of the square?

Solution :

(a)

Refer to the adjacent figure.

m

m

The moment of inertia of spheres A and B about their common 2 diameter AB = mr 2 each. Also the moment of inertia of 5 spheres C and D about an axis passing through their centre and a parallel to AB = . From the parallel axis theorem, the 2 moment of inertia of spheres C and D about diagonal AB is

a m

m

2

(b)

2 2 2 1  a  2 2 2 mr 2 + m (CO ) = mr 2 + m   = mr + ma . 5 5 5 2 2   Hence the moment of inertia (MI) of the system of four spheres about diagonal AB is I AB = MI of C about AB + MI of B about AB + MI of C about AB + MI of D about AB. 2 2 2 1 2 1 = mr 2 + mr 2 + mr 2 + ma 2 + mr 2 + ma 2 5 5 5 2 5 2 2  8r  8 = mr 2 + ma 2 = m  +a2  . 5  5  Moment of inertia of sphere A about side AD = moment of inertia of sphere D about 2 side AD = mr 2 . Using the parallel axis theorem, moment of inertia of sphere C 5 2 about AD = moment of inertia of sphere B about AD = mr 2 + ma 2 . Hence the 5 moment of inertia of the system of four spheres about side AD is IAD = MI of A about AD + MI of D about AD + MI of B about AD + MI of C about AD  8r 2  8 + 2a 2  . = mr 2 + 2ma 2 = m  5  5 

Perpendicular axis Theorem: z

The moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of its moments of inertia about two mutually perpendicular concurrent axes passing through the plane of the lamina : Iz = I x +I y .

O

y

x NOTE Remember that the perpendicular axis theorem does not apply to three dimensional objects but only to two dimensional objects or, lamina.

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Table of moments of inertia: Expression for moment of inertia of bodies of regular shapes about particular axes of rotation:

1.

2.

3. 4.

Shape of body

Axis of Rotation passing through

Circular ring of mass M and radius R

(i)

Circular disc of mass M and radius R

Sphere of mass M and radius R Cylinder of mass M, radius R and length L

Expression for Moment of Inertia

centre, perpendicular to plane of ring (ii) any diameter (iii) any tangent in the plane of ring (iv) any tangent perpendicular to plane of ring (i) centre, perpendicular to plane of disc (ii) any diameter (iii) tangent in the plane of the disc (iv) tangent perpendicular to plane of disc (i) any diameter (ii) any tangent plane (i) own axis (ii)

centre perpendicular to length

(iii) end faces and to length 5. 6.

7.

One dimensional rod of mass M and length L Rectangular lamina of mass M, length L and breadth B

Rectangular block of mass M and dimensions as follows: Length L , breadth B and height H

(i) (ii)

centre of rod and to length one end and to length

(i)

length of lamina and in its plane breadth of lamina and in its plane

(ii)

MR2 (1/2) MR2 (3/2) MR2 2MR2 (1/2) MR2 (1/4) MR2 (5/4) MR2 (3/2) MR2 (2/5) MR2 (7/5) MR2 (1/2) MR2  R 2 L2  M  +   4 12   R 2 L2  M  +  3   4 2 ML /12 ML2/3 MB2/3 ML2/3

(iii) centre of lamina and parallel to length or breadth in its plane

MB 2 ML 2 or 12 12

(iv)

centre of lamina and to its plane

 L2 + B 2  M    12 

(v)

centre of length and to its plane

 L2 B 2  M  +  3   12

(vi)

centre of breadth and to its plane centre of block and parallel to one edge (length of breadth or height)

 L2 B 2  M  +   3 12 

(i)

(ii)

end face and parallel to length or breadth or height of the block

B2 +H 2   H 2 + L2  M   or M   or  12   12   L2 + B 2  M    12  H 2 B2   L2 H 2  M  + +  or M   or 12  12   3  3  B 2 L2  M  + .  3 12 

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Illustration 10.

Solution :

1 MR 2 . 2 What is the moment of inertia of the disc about (i) a diameter (ii) an axis tangential to its circumference? 1 The moment of inertia of the disc about its axis is MR 2 . Let us consider two 2 perpendicular diameters, one along the x-axis and the other along the y-axis. Then , I x = I y (i) According to the theorem of perpendicular axes, the moment of inertia of the disc about an axis passing through its centre and normal to its plane is given by 1 1 or or I x = MR 2 . IC = I x + I y MR 2 = 2I x 2 4 (ii) Since the disc is uniform, its centre of mass coincides with its centre. Therefore, the moment of inertia of the disc about an axis tangential to its circumference is I t = I CM + Md 2 The moment of inertia of a uniform circular disc of mass M and radius R about its axis is

= I CM + MR 2 1 MR 2 + MR 2 2 3 = MR 2 . 2

=

Illustration 11.

Find the moment of inertia of an arc of a ring of radius R, given that the arc subtends an angle θ at the center of the ring and the mass of the arc is m and the axis is along the line joining center of the ring and the midpoint of the arc.

Solution :

We take a small element of the arc between the angular positions φ and φ + dφ, having a mass dm = λ.R dφ, where λ = mass per unit length. Then, the moment of inertia, I=

θ 2

∫ (R sin φ) dm

Β

Ο

2

−θ 2 θ 2

= 2 ∫ ( R 2 sin 2 φ.λR ) d φ

θ

φ

dm P

R Α

C

0

θ 2

= 2λR 3 ∫ sin 2 φ d φ 0

Since, ∴

1 = 2λR 3 × (θ − sin θ) . 4 m λ= Rθ 1 sin θ   . I = mR 2  1 − θ  2 

Note: The moment of inertia about an axis perpendicular to this axis, lying in the plane of the ring, is I1=

1  sin θ  mR 2  1 + . 2 θ  

We obtain this using the perpendicular axis theorem.

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TORQUE The torque due to a force (or, forces) describes the tendency of the force to cause rotation or alter the state of G rotational motion of the body on which it acts. Torque due to a force F acting at A about an axis is given by, G G G τ = rA × F , G where r A is the position vector of the point A where the force F acts on the body. The origin of coordinates is chosen to lie on the axis. Newton’s second law leads to the expression, τ=Iα where I may be treated like a scalar for symmetric bodies and the axis of rotation is parallel to an axis of symmetry. PROBLEM SOLVING TECHNIQUES The general approach to solving complex problems involving rigid body motion is summarised below. (i) (ii) (iii)

Draw a Free Body Diagram (FBD). Select coordinates for describing the motion of each body, and identify the constraints, if any. Include pseudo-forces in the diagram, if required. Apply Newton’s 2nd Law of motion to each body: For rotation, take the net torque due to all the forces, about an axis passing through the centre-of-mass and apply , τnet = I α .

(iv) (v)

In case of pure rotation, it is convenient to take the net torque about the axis of actual rotation instead of the centre of mass. Solve all the equations in steps (ii) and (iii) for the unknown quantities, taking care to perform appropriate checks. Use kinematics, if velocities and displacements are required to be found in the problem.

Illustration 12.

Solution :

Two masses m1 = 15 kg and m2 = 10 kg are attached to the ends of a cord which passes over the pulley of an Atwood’s machine. The mass of the pulley is M = 10 kg and its radius is R = 0.1 m. Calculate the tension in the cord, the acceleration a of the system and the number of revolutions made by the pulley at the end of 2 seconds from the start. There is sufficient friction to prevent slipping between the cord and the pulley. (Take g = 9.8 m/s2)

M

R

T2 m2 m2g

T1 m1 m1g a

The figure shows an Atwood’s machine. As the pulley has a finite mass, the two tensions T1 and T2 are not equal. If a be the linear acceleration of the system, then according to Newton’s laws, we get, m 1 g −T1 = m 1a … (i) T2 − m 2 g = m 2 a … (ii) Net torque on the pulley in the clockwise direction, = (T1 −T 2 ) R … (iii) We know that, Torque = Moment of inertia × Angular acceleration = I ×α

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1 Linear acceleration a MR 2 and α = = . 2 R R

where, I = Torque =

1 a MR 2 × 2 R

… (iv)

From equation (iii) and (iv), we get

(T1 −T2 ) =

1 Ma 2

… (v)

Adding equation (i) and (ii) , we have

( m 1 − m 2 ) g = ( m 1 + m 2 ) a + (T1 −T2 ) . Substituting the value of (T1 −T2 ) from equation (v), we get,

( m1 − m 2 ) g = (m1 + m 2 ) a + ∴

1 Ma 2

( m1 − m 2 ) g

a=

… (vi)

1 m1 + m 2 + M 2

Given that, m 1 = 15 k g , m 2 = 10 k g and M = 10 kg ∴

α=

( 25 − 10 ) × 9.8 15 + 10 + 5

.

From equation (i), T1 = m 1 g − m 1a = m1 ( g − a ) = 15 ( 9.8 − 1.63 ) = 122.55 N . From equation (ii), T2 = m 2 ( g + a ) = 10 × ( 9.8 + 1.63 ) = 114.3 N . Angular acceleration α = 16.3 radian/s2 θ=

1 (16.3)( 2 )2 = 32.6 radians. 2

∴ Number of revolutions = Illustration 13.

32.6 = 5.2. 2π

A uniform cylinder of radius R and mass M can rotate freely about a stationary horizontal axis O (see figure). A thin cord of length A and mass m is wound on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length x of the hanging part of the cord. The wound part of the cord may be supposed to have its centre of gravity on the cylinder-axis.

x

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Solution :

Let m ' be the mass of hanging part of the cord. m m'= x . A Let the cord be descending with an acceleration a. Then m  m  … (i)  xg  −T =  x  a  A   A  τ = TR = I α a MR 2 m + ( A − x ) R 2 and α = where I = A 2 R  MR 2 m a TR =  ∴ + (A − x ) R 2  A  2 R M m  + ( A − x ) a T = A  2  From equation (i) and (ii) we get m  M m  m + (A − x ) a =  x  xg  −  A  A   2   A mxg m M m     or + ( A − x ) a . x +  = A 2  A   A  Solving we get, mxg a= . 1   A m + M  2   or

… (ii)  a 

ANGULAR MOMENTUM Angular momentum of a rigid body undergoing pure rotation about a fixed axis is given by G G L =Iω where I is the moment of inertia of the body about the axis of rotation and ω , its angular velocity. It is related to torque much like linear momentum is related to force: G dpG F = dt G G dL τ= . dt G Angular momentum is a vector quantity and it is directed along ω , provided I can be treated as a scalar ( i.e. the rotation is along one of the axes of symmetry). Angular momentum is conserved if the net torque acting on a body about the given axis is zero. The law of conservation of angular momentum is frequently used, for example in analysing motion under a central force, collisions between extended bodies, etc. Illustration 14.

A symmetrical body is rotating about its axis of symmetry, its moment of inertia about the axis of rotation being 1 kg-m2 and its rate of rotation 2 rev./sec. (a) What is its angular momentum ? (b) What additional work will have to be done to double its rate of rotation ?

Solution :

(a)

As the body is rotating about its axis of symmetry, the angular momentum vector coincides with the axis of rotation. ∴ Angular momentum L = I ω … (i)

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1 2 Iω 2 or 2 E = I ω2 from which, we get,

Kinetic energy of rotation E =

∴ Iω=

(

2IE

)

… (ii)

… (iii) From equation (i) and (ii), L = 2IE ω = 2 rev/sec = 2 × 2π or 4π radian s-1. 1 2 ∴ E = × 1× ( 4 π ) = 8π2 joules. 2

( 2 IE ) =

Now, L = = (b)

Illustration 15.

Solution :

( 2 × 1 × 8π )

(16π ) = 4 π joule 2

2

= 12.57 kg .m 2/s.

When the rate of rotation is doubled, i.e,. 4 rev/sec or 8π radians/sec. the kinetic energy of rotation is given by 1 2 E = × 1 × ( 8π ) = 32π2 joules. 2 Additional work required = Final K.E. of rotation – Initial K.E. of rotation = 32 π2 − 8π2 = 24 π2 = 236.8 joules.

A uniform flat disc of mass M and radius R rotates about a horizontal axis through its centre with angular speed ω0 . (a) What is its kinetic energy? Its angular momentum? (b) A chip of mass m breaks off the edge of the disc at an instant such that the chip rises vertically about the point at which it broke off. How high above the point does it rise before starting to fall? (c) What is the final angular momentum and energy of the disc.

v ω

(a)

The kinetic energy of rotation T R is given by

(b)

1 2 Iω . 2 1 Here, I = MR 2 and ω = ω0 2 1 TR = M R 2 ω02 . 2 The angular momentum, L =Iω 1 or L = M R 2 ω0 . 2 We know that v = rω. v = R ω0 . Here, If the particle rises to a height h, then R 2 ω02 v2 v 2 = 2g h or h = = . 2g 2g If the angular momentum of remainder be LR, then applying the law of conservation of angular momentum we have 1 L R + m R 2 ω0 = MR 2 ω0 2 TR =

(c)

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1  LR =  M − m  R 2 ω02 . 2   Similarly, applying the law of conservation of energy, we get 1 1 (TR )R + mR 2 ω02 = MR 2 ω02 2 4 1  1  (TR ) R =  M − m   R 2 ω2 . 2  2 

ANGULAR IMPULSE: The torque due to an impulsive force when integrated over the time of action of the force gives a quantity known as angular impulse. Mathematically, it is useful in the solution of problems involving impulsive forces G G acting on extended bodies, tending to cause rotation and given as J θ = ∫ τ dt . Illustration 16.

p

Two uniform thin rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02 kg respectively are rigidly jointed, end to end. The combination is pivoted at the lighter end P as shown in the figure such that it can freely rotate about the point P in a vertical plane. A small particle of mass 0.05 kg, moving horizontally strikes the lower end of the combination and sticks to it. What should be the velocity of the particle so that the system just rises to a horizontal position?

A

B m

Solution :

v

p

The situation is shown in figure. Let A be the length of each rod and m A and m B their respective masses. The torque about pivot is zero. Therefore angular momentum of the system is conserved i.e., mv ( 2A ) = I ω … (i) where I is the moment of inertia of the system about point P and ω is the angular velocity of combination just after the collision. Now we shall calculate the value of I.

A l

Bl m

v

Moment of inertia of rod A about P i.e., IA is given by 1 I A = m A A2 3 1 2 = × 0.01 × ( 0.6 ) = 1.2 × 10 −3 kg-m2. 3 The moment of inertia of rod B about P is

`

2

1  3A  m B A 2 + m B   (by the theorem of parallel axes) 12  2 28 28 2 m B A2 = × 0.02 × ( 0.6 ) 12 12 = 1.68 ×10 −2 kg-m2. The moment of inertia of particle about P is I C = m (A + A) 2 = 4m A2 IB =

∴

Moment of inertia I of the system about P is given by

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I = IA + IB + IC = 9 × 10-2 kg – m2

… (ii)

Substituting the value of I in equation (i) we get 0.05v × ( 2 × 0.6 ) = 9 ×10 −2 ω

(

or

ω=

0.05 × ( 2 × 0.6 )

(

9 × 10 −2

)

)

2 ×v = v 3

… (iii)

When the rod moves, the rotational kinetic energy of the system is converted into gravitational potential energy i.e., 1 2 I ω = (m A + m B + m ) g Y CM … (iv) 2 where Y CM is the distance of the centre of mass of system from P. Y CM is given by

or

Y CM =

m A y 1 + m B y 2 + my 3 (m A + m B + m )

Y CM =

0.01 × 0.3 + 0.02 × 0.9 + 0.05 × 1.2 81 = . 0.01 + 0.02 + 0.05 80

Substituting the values in equation (iv), we get 2 ( 0.01 + 0.02 + 0.05) × 9.8 × 81 1 2  × 9 × 10 −2  v  = . 2 80 3  Solving this equation for v, we get, v = 6.3 m/s.

(

)

Illustration 17.

A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of mass m moving a at a speed v perpendicular to the length of the rod strikes it at a distance from the centre and stops 4 after the collision. Find (a) the velocity of the centre of the rod and (b) the angular velocity of the rod after the collision.

Solution :

We take the rod and the particle together as a single system. As there is no resultant force on the system as a whole, the linear momentum of the system will remain constant. Further there is no resultant external torque on the system and so the angular momentum of the system about any point remains constant. Suppose the velocity of the centre of the rod is V and the angular velocity about the centre is ω (see figure) after the collision.

ω

P

a/4

A

a

(a)

A

r0

v

(b)

(a)

The linear momentum before the collision is mv and that after the collision is MV. Thus, m v . mv = MV , or V = M

(b)

Consider the angular momentum of ‘the rod plus the particle’ system about the midpoint of the rod, A. Initially the rod is at rest. The angular momentum of the particle about A is a  L = mv   . 4 After the collision, the particle comes to rest. The angular momentum of the rod about A is, 17

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G G G G r0 ||V , r0 ×V = 0 G G Thus, L = Lcm . Hence the angular momentum of the rod about A is Ma 2 L =Iω= ω. 12 Mva Ma 3mv = ω or, ω = . Thus, Ma 4 12

ROTATION AND TRANSLATION Any general motion of a rigid body involves both translation and rotation. We will not consider all types of general motion of rigid bodies, only pure rolling. A body rolls on a surface - for a example, a ball rolls on the ground, a car tyre rolls on the road etc, - in all these cases (a) (b)

The body undergoes rotation. The point of contact of the body has zero velocity with respect to the surface on which it is undergoing pure rolling.

ROLLING MOTION We will, therefore, define pure rolling as follows: A body is said to undergo pure rolling on a surface if the point of contact of the body has zero relative velocity with respect to the surface on which it is undergoing rolling, and for sustained pure rolling, the relative acceleration along the surface is zero. In addition, the particle at the point of contact of the rolling body moves away from the point of contact as rolling continues.

ω r v P Q

The adjoining figure shows a cylinder rolling on a stationary horizontal surface: v P = v − ωr vQ = 0 v PQ = v P − v Q = 0 or v = ωr = 0 . In addition the tangential relative acceleration. dv PQ dv d ω = a PQ = 0 = − .r dt dt dt for pure rolling to be sustained.

Kinetic Energy in Rolling Motion: The kinetic energy of a body undergoing pure rolling is given by the sum of two terms – (a) (b)

The kinetic energy of translation of the centre of mass. The kinetic energy of rotation about an axis through the centre of mass. 1 1 2 + I cm ω2 Thus, KE = mv cm 2 2 along with the constraint appropriate for rolling with or without slipping. 18

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Illustration 18.

A cylinder is released from rest from the top of an incline of inclination θ and length A . If the cylinder rolls without slipping, what will be its speed when it reaches the bottom?

Solution :

Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it reaches the bottom is v. As the cylinder rolls without slipping, its angular speed about its axis is ω = v / r . The kinetic energy at the bottom will be 1 1 K = I ω2 + mv 2 2 2 11 1 1 1 3  =  mr 2  ω2 + mv 2 = mv 2 + mv 2 = mv 2 . 22 2 4 2 4  This should be equal to the loss of potential energy mg A sin θ Thus,

3 mv 2 = mg A sin θ 4

or,

v =

4 g A sin θ . 3

FRICTIONAL FORCES ACTING DURING PURE ROLLING Since the relative velocity at the point of contact between two rolling surfaces is zero, the only type of friction that may act during pure rolling is static friction. This force acts as a ‘constraint force’ maintaining rolling. Illustration 19.

A sphere of mass m rolls without slipping on an inclined plane of inclination θ . Find the linear acceleration of the sphere and the force of friction acting on it. What should be the minimum coefficient of static friction to support pure rolling?

Solution :

Suppose the radius of the sphere is r. The forces acting on the sphere are shown in figure. They are (a) weight mg, (b) normal force N and (c) friction f.

N

f

Let the linear acceleration of the sphere down the plane mg be a. The equation for the linear motion of the centre of θ mass is mg sin θ − f = ma … (i) As the sphere rolls without slipping, its angular acceleration about the centre is a/r. The equation of rotational motion about the centre of mass is, 2  a  fr =  mr 2    5   r  2 f = ma … (ii) or, 5 From (i) and (ii), 5 2 a = g sin θ and f = m g sin θ . 7 7 The normal force is equal to mg cos θ as there is no acceleration perpendicular to the incline. The maximum friction that can act is, therefore, µ mg cos θ , where µ is the coefficient of static friction. Thus, for pure rolling 2 µ mg cos θ ≥ mg sin θ 7 2 µ ≥ tan θ . or, 7 2 The minimum coefficient of static friction to support pure rolling = tan θ . 7

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Illustration 20.

A force F acts tangentially at the highest point of a sphere of mass m kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of the centre of the sphere.

Solution :

The situation is shown in figure. As the force F rotates the sphere, the point of contact has a tendency to slip towards left so the static friction on the sphere will act towards right. Let r be the radius of the sphere and a be the linear acceleration of the centre of the sphere. The angular a acceleration about the centre of the sphere is α = , as there r is no slipping. For the linear motion of the centre F + f = ma

F f

… (i)

and for the rotational motion about the centre, 2  a  Fr − fr = Ia =  mr 2    5  r  2 F − f = ma … (ii) or, 5 From (i) and (ii), 7 10F 2F = m a or, a= . 7m 5

Rolling on a Horizontal Surface: If a body is undergoing pure rolling on a flat horizontal surface, and no additional forces (or torques) are acting on it except contact forces and gravity (weight), no frictional force acts upon the body. v c = r ω = constant F f = 0 (no friction) On the other hand, if a force F acts on the body the conditions to be satisfied are (i)

For Linear Motion F − Ff = m

(ii)

For Rotation

dv c . dt

dω dt b k2 dω F + Ff = m 2 r or, r r dt and the equation of the constraint v c = r ω . F .b + F f r = I

 b /r − k 2 /r 2  After solving we get, F f = −   2 2  1+ k / r  where k = the radius of gyration and it is assumed that the centre of mass coincides with the geometrical centre of the body.

Rolling on an inclined plane: It is clear from the above discussion that the direction and magnitude of friction are determined by forces and torques acting on the body, other than contact forces.

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If the body rolls on an inclined plane of inclination α , then one can write : dv F + m g sin α − F f = m c dt dω dω F .b + F f r = I = mk 2 dt dt 2 b mk  d ω  F + Ff = 2  r . or, r r  dt  Using the constraint, v c = ωr , we get 2 k2 b  F  2 −  k mg sin α r  r2 r Ff =  + . 2 k k2 1+ 2 1+ 2 r r 2 k 2 If F = 0, F f = r 2 mg sin α , acting backward. k 1+ 2 r Note that this expression does not depend on whether the body rolls up or down the inclined plane. On the other hand if it is not clear in a certain situation as to whether it is the case of pure rolling or not, it is usually advisable to assume that pure rolling occurs. One should apply the rolling constraint and calculate the force of friction, and the normal reaction as well. The magnitude of the force of static friction always satisfies:

F f = µs N where µs is the coefficient of static friction between the two surfaces. FRICTION AND ROLLING (a)

Pure rolling and static friction go hand-in-hand. If pure rolling occurs, there is no relative velocity between the surfaces at the point of contact and therefore, only static friction may act between the bodies. It does not necessarily reach its limiting value. The correct expression is:

F fr ≤ µs N where

F fr

… (i)

represents the force of static friction, N, the normal reaction, and

µs

, the coefficient of static friction between

the surfaces. If it is known that pure rolling occurs,

F fr

and N should be treated as unknown quantities which must be calculated from

Newton’s laws. Equation (i) then provides a consistency check for the problem. (b)

Another question that is frequently asked, quite often at the point where the FBD is being drawn, is the direction of the force of static friction. The answer is: assume it to be an unknown vector i.e. let its component along the surface, along the chosen direction, be F fr . If F fr > 0 as a result of the solution, the choice previously made is correct, otherwise it needs to be reversed.

(c)

If it is known that the surfaces are sliding, friction is kinetic and its direction is such as to oppose relative velocity between the surfaces. Its magnitude is µ k N .

(d)

In problems where it is not clear as to whether friction is static or kinetic, a safe method is to assume that it is static – solve accordingly and check for consistency: If the check fails for a set of surfaces/bodies, assume that

F fr

is kinetic, i.e.

F fr

=

µk N

but along the proper direction

that emerged as a result of the previous solution (static assumption). The process is repeated till the result is found to be consistent. (e)

Are there any shortcuts? No, there aren’t any universal “shortcuts”. A “shortcut” may work in a few simple cases, but it would fail in other cases. It is best to avoid these.

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(f)

There do exist problems which cannot be solved entirely on the basis of Newton’s laws alone, with or without constraints. Most of these problems are outside the purview of the syllabus.

(g)

In case you have been left wondering about rolling friction, we will tell you only this: we have ignored it because it is not so large as to be of much consequence here.

INSTANTANEOUS AXIS OF ROTATION A rigid body that is in motion in general, undergoes a combination of translation and rotation. Rolling motion, for example, is a combination, of translation and rotation. For such motion, the axis of rotation changes in position and /or orientation. The instantaneous velocity of any particle, P, of the rigid body, can be written as: G G G G v P = v 0 + ω× rP G G where v 0 is the velocity of the origin, located on the axis of rotation, and rP is the position vector of the G particle P about O. Here, ω is the angular velocity of rotation of the rigid body about the given axis. It can be proved that, for any moving rigid body, there exists a straight line passing through the body, for which, the velocity of the particles lying on it is zero. This line is known as the instantaneous axis of rotation, because the instantaneous velocity of any particle of the body, may be written in the form: G G G v P = ω× rP G where rP is the position vector of this particle with respect to a point on this line. This line, therefore, can be regarded as an axis of rotation; but only for an instant. Illustration 21.

A cylinder of radius r undergoes pure rolling on a flat horizontal surface with a speed v, as shown in the figure. Take the x-axis in the direction of motion of the cylinder, the y-axis vertically upward and the z-axis out of the plane of the paper. Find the instantaneous axis of rotation.

Solution :

The angular velocity of rotation of the cylinder is ω = v / r , clockwise about an axis passing through the centre of mass of the cylinder.

y

P x v

O The velocity of any particle, P, is given by : G G g v P = v iˆ + ω −kˆ × rP JJJG G where r P = OP G  v G  JJJG G i.e. v P = v i −  k  ×OP . r  Let Q be the point through which the axis of rotation passes. Thenv Q = 0 (by definition). Using the expression given above, G v G JJJJG G v Q = 0 = v i − k ×O Q r G G JJJJG or, ri = k ×OQ JJJJG JJJJG G ∴ |OQ | = r , and OQ = −r j ;

( )

which is the position vector of the particle in contact with the ground. The line of contact between the cylinder and the flat surface on which it rolls serves as the axis of rotation at any instant.

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ANGULAR MOMENTUM, TORQUE AND KINETIC ENERGY For a rigid body undergoing translation as well as rotation, the total angular momentum can be written as the sum of two parts: (a)

(b)

The angular momentum associated with centre of mass motion, i.e. the angular momentum obtained by regarding the body as a single particle located at its centre of mass (orbital angular momentum). The angular momentum associated with rotation of the body about an axis passing through the centre of mass (spin). G G G G Laxis = rc m × p cm + Lc m ,axis If we apply the result stated above to the body moving as show in the diagram, G G G L nˆ = rcm × mv c m + ( I cm ω ) nˆ '

nˆ

O

nˆ′

r cm

axis

v cm mass = m M.I. = Icm

where nˆ , nˆ ' , are parallel unit vectors in the same direction; the vector nˆ ' passing through the centre of mass.

ROTATIONAL WORK AND POWER: For a rigid body undergoing translation as well as rotation, the work done by all forces is given by JJG G … (i) W net = ∑ ∫ F i .d r i i

JJG G G where, each force F i acts at the point with position vector ri and d r i represents the infinitesimal G displacement of the point of application of the ith force (i.e. the change in ri ). The motion of the rigid body can be studied equally well by studying the motion in terms of the translation of the centre of mass and rotation about the centre of mass. The velocity of any particle i is given by the well known relation: G JG JJG JJG v i = v cm + ω × R i … (ii) JJG where R i represents the position vector of the ith particle with respect to the centre of mass. The expression (i) can be simplified as follows: JJG G dW net = ∑ F .d r i i

G G JG JJG G G dri = ∑ F v cm + ω× R i dt, since v i = dt i JJG G JJG JJG JJG = ∑ F .v cm dt + ∑ F i .(ω × R i ) dt

(

i

)

i

JJG JJG JJG JJG JJG = F net . d R cm + ∑ R i × F i .ω d t . i

(

)

Integrating this we get, JJG JJG JG JJG G G W net = ∫ F net . d R cm + ∫ τ net .d θ , where d θ = ωd t . JJG JJG JG JJG JJG Here F net = ∑ F i is the net force acting on the body, and τ net = ∑ R i × F i , represents the net torque acting i

i

on the body about the centre-of-mass. The first term alone does not give the total work done by all the forces acting on the body. 23 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016

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Illustration 22.

Solution :

F

A uniform solid cylinder of mass m and radius r undergoes pure rolling on a horizontal plane under the action of a horizontal force F pulling it at its topmost point as shown in the adjoining diagram. If the cylinder is displaced by x, what is the work done by F? Find the work done by friction as well.

x

The work done in translation is, JJG JJG ∫ F net d R c m = (F − F fr ).x

… (i)

The work done during rotation is ∫ ( F + Ffr ).r .d θ

(F (F

or

+ F fr ).∫ rd θ

or + F fr ) .x The net work done is: W net = W tran +W rot

… (ii)

= ( F − F fr ) x + ( F + F fr ) x = 2 F .x = F . ( 2x )

The net work done by friction is zero, as static friction is a constraint force. The net work done by all forces is F .( 2 x ) . Illustration 23.

Solution :

Find the angular momentum of a solid cylinder undergoing both translation and rotation, about the point A, as shown in the figure. The mass of the cylinder is m, its radius is r, and it is not necessarily undergoing pure rolling.

O v P

ω

Horizontal surface

A

L A = mv .OP + L cm 1 mr 2 ω 2 (the plus sign is due to the fact that both terms are clockwise). A similar expression applies for Torque. The torque is given by (referring to previous figure): G G G G G Taxis = rcm × F net + ∑ ρi × F i = mvr +

i

G where ρi is the position vector of each particle relative to the centre-of-mass of the rigid body. Illustration 24.

Referring to the previous illustration, find the kinetic energy of the cylinder.

Solution :

K .E =

1 11  mv 2 +  mr 2  ω2 2 22  1 1 mv 2 + mr 2 ω2 . 2 4

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SOLVED SUBJECTIVE PROBLEMS Problem 1.

A particle of mass m is projected at time t = 0 from a point P with a speed v 0 at an angle of 45º to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time t.

Solution:

Let us take the origin at P, x-axis along the horizontal and y-axis along the vertically upward direction as shown in figure.

Y

For horizontal motion during the time 0 to t, v x = v 0 cos 450 = and

x =v xt =

v 0t 2

v

r

v0 2

x

P

.

For vertical motion, v y = v 0 sin 450 − gt = and

y = (v 0 sin 450 )t − =

v0 2

− gt

1 2 gt 2

v0

1 t − gt 2 . 2 2

The angular momentum of the particle at time t about the origin is

(

) (

L = r × p = xiˆ + yjˆ × m v x iˆ + v y jˆ =−

mv 0 2 2

)

gt 2 kˆ

where. iˆ × jˆ = kˆ , which is out of the paper. mv 0

gt 2 in the negative z-direction i.e., 2 2 perpendicular to the plane of motion, going into the plane. Thus, the angular momentum of the particle is

Problem 2.

Solution:

A small solid sphere of mass 1 kg and radius 0.2 metre rolls down along a track shown in figure, without slipping. Find the height h above the base, from which it has to start rolling down the incline, such that the sphere just completes the vertical circular loop of radius 1 metre.

A r R

h

The situation is shown in figure. Let m and r be the mass and radius of solid sphere. When the sphere reaches the point A , it descends through a vertical distance ( h − 2 R + r ) , where R is the radius of circular loop. At A , the loss of potential energy = mg ( h − 2R + r )

…(i)

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Total kinetic energy of the sphere at A = K.E. of the rotation + K.E. of translation of sphere 1 2 1 I ω + mv 2 2 2 2 12 v  1 =  mr 2   2  + mv 2 25  r  2 =

=

7 mv 2 10

… (ii)

For just completing the vertical circular loop mv 2 = mg (R − r ) or

v 2 = (R − r ) g . 7 mv 2 10 7 = m (R − r ) g . 10

Now total kinetic energy of sphere at A =

Loss of potential energy = gain of K.E. mg ( h − 2R + r ) =

7 m (R − r ) g . 10

Solving we get, 17   27 h =  ×R − r  10   10 substituting the given values, we get 17  27  h =  × 1 − × 0.2  = 1.36 m . 10  10  Problem 3.

Calculate the kinetic energy of a tractor crawler belt (see figure) of mass m if the tractor moves with velocity v. There is no slipping. The dimensions of the wheels are as shown in the figure.

A

v D

Solution:

B

C

The velocity of the centre of mass of the tractor wheel is v . velocity of the lower part of the belt in contact with the ground is = v −v = 0 . Velocity of the upper part between the two wheels is = v +v = 2v Suppose that the mass of the upper portion of the belt AB between the two wheels is m1 and the total mass of the portion of the belt in contact with the wheels between BC and DA is m2. The total mass of the belt is:

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The kinetic energy (KE) of the belt = KE of AB + KE of CD + KE of BC + KE of DA =

1 m 1 ( 2v 2

)

2

1 1 v  + 0 + m 2v 2 + m 2 R 2 .   2 2 R 

2

= 2m 1v 2 + m 2v 2 = mv 2 . Problem 4.

A cylinder of mass m is suspended through two strings wrapped around it as shown in figure. Find (a) the tension T in the string and (b) the speed of the cylinder as it falls through a distance h.

T

T mg

Solution:

The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. Suppose the centre of the cylinder falls with an acceleration a. The angular a acceleration of the cylinder about its axis is α = , as the cylinder does not slip over the r strings. The equation of motion for the centre of mass of the cylinder is mg − 2T = ma

… (i)

and for the motion about the centre of mass, it is

or,

1  1 2Tr =  mr 2 α  = mra 2  2 1 2T = m a 2

… (ii)

From (i) and (ii), a=

mg 2 . g and T = 6 3

As the centre of the cylinder starts moving from rest, the velocity after it has fallen through a distance h is given by 2  v2 = 2  g  h 3  4 gh . 3 A solid spherical rotor of radius 30 cm and weight W is connected to a link AB and freely rotates about B with angular velocity 60 radian/second. If the rotor is suddenly allowed to rest on a vertical wall, what time will elapse before it comes to rest? The coefficient of friction between the wall and the rotor is 0.25 and inclination of link with respect to the vertical is 15º.

Problem 5.

v =

A T

15 o

or,

B

r

C

R

r C B W +µR 27

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Solution:

In the figure, R is the normal reaction at C and T is the tension in AB . The rotor has no motion whether horizontal or vertical. Here T cos75o = R and T sin 75o =W + µR W + µR W + 0.25R = R R Solving, we get R = 0.29W ∴

tan 750 =

Moment of inertia, I =

2 2W 2 Mr 2 = r 5 5 g

torque Moment of inertia (µR ) r 5µ gR = Retardation = 2 W 2 2r W . .r 5 g Angular acceleration, a =

where ( µR ) r

= Frictional torque

5 × 0.25 × 9.8 × 0.29W 2 × 0.3 ×W = 5.92 rad/sec2. Time taken by the rotor to come to rest ω = ω0 + αt or, 0 = 60 − 5.92t t = 10.13 sec. So, retardation =

Problem 6.

Solution:

A round cone with half-angle α = 30º and the radius of the base R = 10 cm rolls uniformly without slipping over a horizontal plane. The apex is hinged at the point O (see figure) which is at the same level as the point C, the centre of the base. The velocity of the point C is v = 10 ms −1 . Find (a) the angular velocity vector of the cone and the angle it forms with the vertical and (b) the angular acceleration vector. (a)

O C

There are two angular motions of the cone; one in the horizontal plane with angular v where R cot α = radius of the circle in which C moves. The velocity ω1 = R cot α direction of this vector is upward. The other angular velocity ω2 is about its own axis. Since it rolls without slipping, v = ω2 R ⇒ ω2 =

v . The direction of this vector is R

horizontal and towards O . v2 v2 v + tan 2 α = tan α R R R Let θ be the angle made by the resultant with the vertical. Then ω tan θ = 2 = cot α = cot 300 = 3 ω1 ∴

ωresult = ω12 + ω22 =

∴ θ = 600 . (b)

There is no change in ω1 either in magnitude or direction and hence, the angular acceleration corresponding to ω1 is zero. There is no change in the magnitude of ω2 but there is change in its direction. ∆ θ = angle through which the axis OC rotates in the horizontal plane ω1 ∆t .

ω ω2 +

∆θ

2

ω2∆θ

∆θ ω2

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∴ change in angular velocity = ω2 ∆θ = ω2 ω1 ∆t

Problem 7.

ω1ω2 ∆ t = ω1 ω2 ∆t

∴

α2 =

∴

α res = ω1ω2 =

v2 v v = 2 tan α . R cot α R R

Two horizontal discs rotate freely about a common vertical axis passing through their centres. The moments of inertia of the discs relative to there axis are equal to I1 and I 2 , and their angular velocities are ω1 and ω2 . The upper disc falls on the lower disc and after some time, both discs begin to rotate as a single body due to friction. Find

Solution:

(a)

the steady state angular velocity of the discs, and

(b)

the work performed by the frictional forces in the process.

(a)

Since there is no external torque on the system consisting of both the discs, angular momentum of the system is conserved. I 1ω1 + I 2 ω2 = ( I 1 + I 2 ) ωf ⇒

(b)

ωf =

I 1ω1 + I 2 ω2 I1 + I 2

Work done by frictional forces = change in kinetic energy of the system W fric tio n = change in kinetic energy = Tf −Ti W fr =

or,

1 1 1 ( I 1 + I 2 ) ωf 2 −  I 1ω12 + I 2 ω22  2 2 2 

(I ω + I ω ) 1 1 = ( I 1 + I 2 ) 1 1 2 22 − ( I 1 ω12 + I 2 ω22 ) . 2 2 (I 1 + I 2 ) 2

1 ( I 1ω1 + I 2 ω2 ) 1 − ( I 1ω12 + I 2 ω22 ) 2 2 I1 + I 2 2

W fr =

=−

I 1I 2 ( ω1 − ω2 ) 2 . 2(I 1 + I 2 )

which is very similar to the expression for a perfectly inelastic collision. Problem 8.

A uniform sphere of mass m and radius R starts rolling without slipping down an inclined plane. Find the time dependence of the angular momentum of the sphere relative to the initial point of contact. How will the result be affected in the case of a perfectly smooth inclined plane? Here, θ = angle of inclination of the plane.

Solution:

Let O be the initial point of contact.

N

For linear motion , we write, mg sin θ − f fr = mac m . For rotation,

∴ or,

sinθ mg v cm

a τ = f fr R = I α , where α = c m R Iα mg sin θ − = mac m R Ia mg sin θ − c 2m = mac m R

ffr

O

θ mg

mg cosθ

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or, or,

I  mg sin θ =  m + 2 R  mg sin θt v cm = . I m+ 2 R

I  dv c m    ac m =  m + 2  R    dt

Using the expression for angular momentum ( L axis = L spin + L orbital ) Iv c m + mv cm R R I  mg sin θt  I   L0 = v cm R  m + 2  = R m + 2  I R  R    m+ 2 R = ( mg sin θ ) Rt .

L 0 = Lc m + mv c m R = I ω + mv c m R =

If the plane is perfectly smooth, there is no rotation. ∴ Problem 9.

Solution:

L0 = mv cm R = mg sin θRt .

A truck, initially at rest with a solid cylindrical paper roll, moves forward with a constant acceleration a. The cylinder roll is lying parallel to the forward edge of the truck at a distance d from the rear edge of the truck. Find the distance s which the truck travels before the paper roll moves off the edge of its horizontal surface. Friction is sufficient to prevent slipping between the paper and the truck.

Let a r be the acceleration of the cylinder relative to truck. The different forces acting on the cylinder are shown in fig. Here, ma − f = ma r

d

α f

ar ma

… (i)

1 mr 2 α 2 ar = r α f .r =

… (ii) … (iii)

From equation (i), 2a a  or, or , a −  r  = ar ar = 3  2 1  2a  1 2  d =  t 2 Now ∵s = ut + at  2 3  2   where t is the time required to cover the distance d.  3d  ∴ t =    a  The distance traveled by the truck relative to ground before the paper rolls off is given by 1 1  3d  3d . s = at 2 = a  = 2 2 2  a  30 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016

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Problem 10.

A uniform sphere of mass m and radius r rolls without slipping down from the top of a larger fixed sphere of radius R. Find the angular momentum acquired by smaller the sphere about the centre of the larger sphere as well as its K.E. at the moment when it breaks off from the larger sphere. The initial velocity of the sphere is negligible.

Solution :

The centre of mass of the sphere moves in a vertical circle of radius ( R + r ) . Considering the dynamics of circular motion of the centre of mass of the sphere mg cos θ − N =

m

θ

mv 2 R +r

when it breaks off N = 0 At the point where the smaller sphere breaks off, mg cos θ =

m ω2 r 2 . R +r

(∵v

= ωr )

Since friction is static and gravity is conservative (other forces being constraint forces), mechanical energy is conserved.

or or

1 1 mg ( R + r ) = mg ( R + r ) cos θ + mv 2 + I ω2 2 2 7 m g ( R + r ) = mg ( R + r ) cos θ + m ω2 r 2 10 7 m g ( R + r ) = m ω2 r 2 + m ω2 r 2 10

10 ( R + r ) g . 17 r2 The angular momentum of the sphere is L = L orbital + I ω

or

ω=

2 = mv ( R + r ) + mr 2 ω 5 2 = m ωr ( R + r ) + m ωr 2 5 7   = m ωr  R + r  5   10 7   g (R + r )  R + r  . 17 5   Kinetic energy of the sphere is 1 1 K.E. = mv 2 + I ω2 2 2 1 12 = mr 2 ω2 + mr 2w 2 2 25 7 7 10 ( R + r ) = mr 2 ω2 = mr 2 . g 10 10 17 r2 7 = mg( R + r ) . 17 =m

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SOLVED OBJECTIVE PROBLEMS Problems 1.

A solid body rotates with deceleration about a stationary axis with an angular deceleration α = k ω ; where k is a constant and ω is the angular velocity of the body. If the initial angular velocity is ω0 , then mean angular velocity of the body averaged over the whole time of rotation is ω0 (b) (a) ω0 2 (c)

Solution :

−

ω0 3

ω0 4

(d)

dω = k ω1/ 2 dt t dω = k ∫ dt 1/ 2 ω0 ω 0 ω

⇒

−∫

⇒

−2  ω  = kt ω0

⇒

2 ω0 − 2 ω = kt

⇒

2 ω = 2 ω0 − kt ;

ω

1   ω =  ω0 − kt  2  

2

The body comes to a stop ( ω = 0 ) in time t0 =

2 ω0 k

kt   Further ω =  ω0 −  2   ⇒

ω = ω0 + t0

Since ωav =

∫0 t ∫0

k 2t 2 − kt ω0 4

ωdt 0

2

dt

=

ω0t 0 +

t2 k 2 t 03 − k ω0 0 4 3 2 t0

⇒

ωav = ω0 +

2 t k 2 t0 − k ω0 . 0 4 3 2

⇒

ωav = ω0 +

k 2 4ω0 k ω0 2 ω0 − 12 k 2 2k

⇒

ωav = ω0 +

ω0 − ω0 3

⇒

ωav

Ans.

=

ω0 . 3

(c)

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Problems 2.

A uniform wheel of moment of inertia I is pivoted on a horizontal axis through its centre so that its plane is vertical as shown in the figure. A small mass m is stuck on the rim of the wheel as shown. The angular acceleration of the wheel when mass is at point A is α A and when mass is at point B is α B . Then αB αB = sin θ (b) = sin 2 θ (a) αA αA (c)

Solution :

αB = cos θ αA

b

αB = cos2 θ αA

(d)

(c)

A centrifuge consists of four solid cylindrical containers, each of mass m at radial distance r from the axis of rotation. A time t is required to bring the centrifuge to an angular velocity ω from rest under a constant torque τ applied to the shaft. The radius of each container is a and the mass of the shaft and the supporting arms is small compared to m. Then

(a) t =

(c)

4mr 2 ω τ

t=

(b)

2ma 2 ω t= τ

t=

(d)

r

a

τ

m

2m (a 2 + 2r 2 )ω τ

(

)

2m r 2 + 2a2 ω τ

L = τt

⇒

Problems 4.

A

τB I α B b cos θ = = = cos θ . b τA I α A

Ans.

Solution :

θ

The torque due to the weight of m at A is τA and that when it is at B is τB ⇒

Problems 3.

B

1   4  mr 2 + ma 2  ω 2   t = τ

⇒

t =

Ans.

(b)

2m ( a 2 + 2r 2 ) ω τ

.

One quarter sector is cut from a uniform disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is 1 1 MR 2 (b) MR 2 (a) 2 4 (c)

1 MR 2 8

R

R

2 MR 2 .

(d)

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Solution :

If we assume complete disc to be present, then it would have a mass 4 times the mass of the sector. Then, moment of inertia of the complete disc is I disc =

Problems 5.

Now,

I sector =

I disc 4

⇒

I sector =

1 MR 2 . 2

Ans. (a) Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to (a) I (b) I sin 2 θ I cos2 θ

(c) Solution :

1 1 M disc R 2 = ( 4 M sector ) R 2 2 2

θ (d) I cos2   2

I AB = I A ' B ' = I and I CD = I C ' D '

A

If I 0 be the moment of inertia of the square plate about an axis passing through O and perpendicular to the plate, then by the perpendicular axis theorem

or

I 0 = I AB + I A ' B ' = 2I AB

.... (1)

I 0 = I CD = I C ' D ' = 2I CD

.... (2)

C′

D

θ θ

A′

B′ D′

C

From ((1) and (2))

B

I CD = I AB = I . Ans. Problems 6.

A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in the figure. The magnitude of angular momentum of the disc about the origin O is 1 (a)   MR 2 ω (b) MR 2 ω 2 (c)

Solution :

(a)

3 2   MR ω 2

ω M

R

O

(d) 2MR 2 ω

The angular momentum of a body L may be expressed as the sum of two parts,

(

(a) one arising from the motion of the centre of mass of the body L orbital

) (

(b) the other from the motion of the body with respect to its centre of mass L spin

)

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⇒ Lto tal = LC .M . + M ( rC .M ×v C .M

)

For this problem LC .M = I ω =

1 MR 2 ω and 2

M ( rC .M ×v C . M ) = MRv CM = MR ( R ω )

(

)

⇒ M rC .M ×VC .M = MR 2 ω ⇒ Lto tal = Ans. Problems 7.

(c)

A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown in the figure. The moment of inertia of the loop about the axis XX' is ρL3 ρL3 (b) (a) 8π 2 16 π2 (c)

Solution :

1 3 M R 2 ω + MR 2 ω = MR 2 ω 2 2

5ρL3 16 π2

X

|

X

0

90

O

3ρL3 8π2

(d)

Mass of the loop = M = L ρ Further if r is the radius of the loop, then 2 πr = L L ⇒ r = 2π 3 Moment of inertia about XX' is I = M r 2 2 3 2 L 3ρL 3 = . ⇒ I = (L ρ ) 2 2 8π2 2π

(

Ans. Problems 8.

(d)

A uniform solid cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to X-Y plane) with its axis along the Y-axis and free to roll on the platform. The platform is given a motion in X-direction given by x = A cos ωt . There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder as measured about its centre of mass 1 MRAω2 (b) MRAω2 (a) 2 (c)

Solution :

)

2mRAω2

τ=Iα=

mRωA 2 cos 2 ωt

(d)

1  linear acceleration  MR 2 ×   2 R  

acceleration = −ω2 A sin ωt or ( linear acceleration )max = ω2 A ∴

( τ)max =

Ans.

(a)

1 MR 2 ω2 1 = MRA ω2 . 2 R 2 35

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Problems 9.

A metal ball of mass m is put at the point A of a loop track and the vertical distance of A from the lower most point of track is 8 times the radius R of the circular part. The linear velocity of ball when it rolls of the point B to a height R in the circular track will be (a)

[10gR ]

(c)

 7 gR   5   

8R

B R

1/ 2

 gR  (b) 7    10 

1/ 2

1/ 2

Solution :

A

[5gR ]

1/ 2

(d)

Applying the conservation of energy at points A and B, we have mg ( 8R ) =

mg ( 8R ) =

or =

mv 2 1 2 + I ω + mgR 2 2 2

mv 2 1  2  v  +  mR 2   + mgR 2 25  R 

7 mv 2 + mgR 10

mg ( 8R − R ) =

7 mv 2 10

[10gR ]1/ 2 . Ans. Problems 10.

A uniform solid cylinder of mass m can rotate freely about its axis which is kept horizontal. A particle of mass m 0 hangs from the end of a light string wound round the cylinder. When the system is allowed to move, the acceleration with which the particle descends is 2m 0 g 2m 0 g (b) (a) m 0 + 2m m + 2m 0 (c)

Solution :

(a)

m0g m + m0

2mg m + 2m 0

(d)

Suppose that the tension in the string is T. Then,

= m 0 g −T = m 0 a

∴

T = m0 (g −a )

where

T a

a = acceleration

m.g

Further, T .r = I α α = angular acceleration and T.r = moment of force acting on cylinder or T =

Iα r

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 mr 2   α  mr α ma = =   = 2 2  2  r  2T 2 m 0 ( g − a ) = m m Solving for a, we get  2m 0 g  a = .  m + 2m 0  ∴

Problems 11.

a=

Ans. (b) The torque τ on a body about a given point is found to be A × L where A is a constant vector and L is angular momentum of the body about that point. From this it follows that dL (a) is perpendicular to L at all times. dt (b) the component of L in the direction of A does not change with time. (c) the magnitude of L does not change with time. (d) L does not change with time.

Solution :

Due to law of conservation of angular momentum, L = constant i.e. L .L = constant or,

d L .L = 0 dt

(

)

or, 2 L .

dL =0 dt

or, L ⊥

dL . dt

Since τ = A × L dL = A ×L dt i.e.

dL must be perpendicular to A as well as L . dt

Further the component of L along A is

(

)

d dL dA +L. =0 A .L = A . dt dt dt

A .L . Also A dL dA   and = 0 ∵ A ⊥ dt dt  

or, A . L = constant i.e.

A .L = x = constant . A

dL ( or τ ) is perpendicular to L , hence it cannot change magnitude of L but can dt surely change direction of L . Ans. (a), (b), (c) Since

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Problems 12.

A particle of mass m is projected with velocity v making an angle of 450 with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is mv 3 (a) zero (b) 4 2g (c)

Solution :

mv 3

(d) m 2 gh 3

2g

L = mv x h ⇒

L = m (v cos 45° )

⇒

L=

v 2 sin 2 45° 2g

mv 3 4 2g

Further L = m v x h

Problems 13.

Solution :

⇒

L = m (v cos 45° ) h o

But h =

v 2 sin 2 450 2g

⇒

v = 2 gh

⇒

L=

⇒

L = m 2 gh 3 .

Ans.

(b) (d)

mv 2

2 gh h

A constant force F is applied at the top of a ring as shown in figure. Mass of the ring is M and radius is R. Angular momentum of particle about point of contact at time t (a) is constant

(b) increases linearly with time

(c) is 2F R t

(d) decreases linearly with time

F

From angular impulse = change in angular momentum we have L = τ t or L = F(2R ) t i.e., L varies linearly with time.

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Problems 14.

A spherical body of radius R rolls on a horizontal surface with linear velocity v. Let L1 and L2 be the magnitudes of angular momenta of the body about centre of mass and point of contact P. Then

v P

Solution :

(a) L2 = 2L1 if radius of gyration K = R

(b) L2 = 2L1 for all cases

(c) L2 > 2L1 if radius of gyration K < R

(d) L2 > 2L1 if radius of gyration K > R

L1 = Iω = MK 2 ω

… (1)

L 2 = Iω + MRv

= MK 2 ω + MR (ωR )

(as v = Rω )

= Mω ( K 2 + R 2 )

… (2)

From equations (1) and (2), we can see that L2 = 2 L1 when K = R and L2 > 2L1 when K > R Problems 15.

A hollow sphere of radius R rests on a horizontal surface of finite coefficient of friction. A point object of mass m moved horizontally and hits the sphere at a height of R/2 above its center. The collision is instantaneous and completely inelastic. Which of the following is/are correct ? (a) total linear momentum of the system is not conserved (b) total angular momentum about center of mass of the system remains conserved (c) the space gets finite angular velocity immediately after collision (d) The sphere moves with finite speed immediately after collision

Solution :

Impulse due to normal reaction is finite. So friction force gives finite impulse. Therefore frictional torque causes a finite angular impulse about center of mass of system so angular momentum about center of mass of system will change. All other options are correct.

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PASSAGE BASED PROBLEMS # Write-up I [Questions 1 to 3] A straight is wrapped several times on a cylinder of mass M and radius R. the cylinder is pivoted about its axis of symmetry. A block of mass m tied to the string rests on a support so that the string is slack. the block is lifted upto a height h and the support is removed. (shown in figure)

R

m m

Problems 1.

What is the angular velocity of cylinder just before the string becomes taut (a) zero

gh R

(c) Problems 2.

(c)

Solution :

(b)

2gh R

(d)

2 gh R

When the string experience a jerk, a large impulsive force is generated for a short duration, so that contribution of weight mg can be neglected during this duration. Then what will be speed of block m, just after string has become taut (a)

Problems 3.

h

2gh  M 1 + m  gh  M 1 + m 

2gh M  1 + 2m 

(b)

gh M  1 + 2m 

(d)

If M = m, what fraction of KE is lost due to the jerk developed in the string (a)

1 2

(b)

2 3

(c)

1 3

(d)

1 4

(1) Just before the string becomes taut, the block falls freely, so v 0 = 2gh. There is no tension in the string, so nothing causes the cylinder to spin, so ω 0 = 0. The kinetic energy of the system is K 0 =

1 1 mv02 + Iω 02 = mgh 2 2

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(2) When the string experiences a jerk, the large impulse developed is of very short duration so that the contribution of weight mg can be neglected during this time interval. The angular momentum of the system is conserved, as the tension is internal force for the system. Thus we have

Li = L f 1 mv1R + MR 2 ω1 = mv 0 R = m 2ghR 2 The string is inextensible, so v1 = Rω1. On solving for ω1 we get

ω1 =

2gh

R 1 + ( M / 2m) 

v1 = Rω1 =

2gh 1 + ( M / 2m) 

The final kinetic energy K1 is given by

1 1 K1 = mv12 + Iω12 2 2 2 1 11  v  = mv12 +  MR 2   12  R  2 2 2

1 M =  m +  v12 2 2 =

 mv02 1   2 1 + ( M / 2m) 

=

K0 1 + ( M / 2m)

(3) For M = m,K1 =

2K 0 , so the fraction lost is 3

 K 0 − K1  1  K  = 3 0

# Write-up I [Questions 4 to 6] A man of mass 100 kg stands to rim of a turn table of radius 2m, moment of inertia 4000 kg. the table is mounted on a vertical smooth axis, through its center. The whole system is initially at rest. The man now walks on table with a velocity 1m/s relative to earth Problems 4.

With what angular velocity will the turn table rotate (a) 0.5 rad/sec (b) 0.1 rad/sec (c) 0.05 rad/sec

Problems 5.

(d) 0.2 rad/sec

Through what angle will the turn table have rotated when the man reaches his initial position on it (a)

π rad/sec 11

(b)

3π rad/sec 11

(c)

2π rad/sec 11

(d)

4π rad/sec 11 41

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Problems 6.

Solution :

Through what angle will it have rotated when the man reaches his initial position relative to earth (a)

π rad/sec 5

(b)

2π rad/sec 5

(c)

2π rad/sec 11

(d)

π rad/sec 11

(1) By conservation of angular momentum on the man-table system,

L1 = L1 or 0 + 0 = I m ω m + I t ωt ωt = −

Imω m v 1 where ω m = = rad / s I1 r 2

ω t = −100 ( 2) × 2

=−

1/ 2 4000

1 rad / s 20

Thus the table rotates clockwise (opposite to man) with angular velocity 0.05 rad/s. (2) If the man completes one revolution relative to the table then

θ mt = 2π;2π = θ m − θ t 2π = ω m t − ω t t (where t is the time taken) t = 2π / ( ω m − ω t ) = 2π ( 0.5 + 0.05) Angular displacement of table is

θ t = ω t t = −5.05 × ( 2π / 0.55) =−

2π radian 11

The table rotates through 2π /11 radians clockwise (3) If the man completes one revolution relative to the earth, then

θ m = 2π time =

2π 2π = ω m 0.5

During this time, angular displacement o the table,

θ t = ω t ( time) = −0.05 ×

2π 0.5

π θ t = − radian 5 θ t = 36° in clockwise direction

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MATCHING TYPE PROBLEM Problems 7.

A wheel of radius R is rolling on the horizontal table. The speed of upper and lower point is v1 and v2 in the direction of motion with respect to table. The following table is given on this fact.

A

v1

B v2 C

Solution :

v1 + v 2 2R

(A)

The speed of centre of mass of wheel

(B)

The angular speed of the Wheel

(Q)

v1 − v 2 2R

(C)

The speed of extreme point B on the wheel

(R)

v1 + v 2 2

(D)

The condition for pure rolling

(S)

v1 − v 2 2

(P)

(T)

v12 + v 22 2

(U)

v12 − v 22 2

(V)

v1 = 0

(W)

v2 = 0

V1 = Vcm + ωR V2 = Vcm − ωR VB = ωR 2 Ans. (A) → (R), (B) → (Q), (C) → (T), (D) → (W)

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ASSERTION-REASON TYPE PROBLEMS Choose any one of the following responses : (A) (B) (C) (D)

If Assertion is true and Reason is a correct explanation of Assertion. If Assertion is true and Reason is not correct explanation of Assertion. If Assertion is true but Reason is false. If Assertion is false and Reason is true

Problems 8.

Solution :

Assertion : Radius of gyration depends on axis of rotation. Reason : Radius of gyration is rms distance of particles of the body from the axis of rotation. (a) A

(b) B

(c) B

(d) D

K=

r12 + r22 ....... + rnn n

Ans. (a) Problems 9.

Assertion : there are two propellers in a helicopter. Reason : Angular momentum is conserved. (a) A

(b) B

(c) B

(d) D

Ans. (a) Problems 10.

Assertion : A wheel moving down a frictionless inclined plane shall undergo slipping (not rolling). Reason : For rolling, torque is required, which is provided by tangential frictional force. (a) A

(b) B

(c) B

(d) D

Ans. (a)

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SECTION–I SUBJECTIVE TYPE QUESTIONS LEVEL – I 1.

A force F = a + bx acts on a particle in x direction where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.

2.

A force F = Aˆi + Bjˆ acts at a point whose radius vector w.r.t origin is r = aiˆ + bjˆ, where a,b,c, A, B are

G

G

G constants and ˆi, ˆj are unit vectors along X and Y axes. Calculate the torque τ and the arm A of the force G F w.r.t. point O.

3.

A meter stick is held vertically with one end on the floor and is then allowed to fall. find the velocity of the other end when it hits the floor, assuming that the end on the floor does not slip.

4.

A wheel of radius 0.50 m initially at east and reaches a constant angular acceleration of 3.0 radian/sec2. Calculate (i) The angular displacement and the angular speed of wheel 2.0 second later, (ii) The tangential speed, the tangential acceleration, the centripetal acceleration and the resultant acceleration of the particle on the rim at end of 2.0 sec.

5.

A diatomic molecules is supposed to be consisted of two masses m1 and m2 separated by a fixed distance r. Determine an expression for the moment of inertia of the molecule about an axis passing through center of mass and perpendicular to bond length.

r1 A

m1

(r2 = r – r1)

C

m2

B

Hence determine the moment of inertia of HCl molecular about an axis passing through its center of mass and perpendicular to the bond. Given : inter nuclear distance 4 = 1.3 Å, mass of proton = 1.7 × 10-27kg, atomic weight of chlorine = 35. 6.

The flat surface of hemisphere of radius R is fixed to one flat surface of a cylinder of radius R and length A . Both are of same material. Let the total mass be M. Prove that the moment of inertia of the combination about the axis of the cylinder of is given by

Y

A R

 A 4R  MR  +  2 15  2R A+ 3 2

O

R

X

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7.

Two masses m1 = 12kg and m2 = 8kg are tied to the ends of a string which passes over a pulley of an Atwood’s machine. the mass of the pulley is M = 10kg and its radius R = 0.1m. Find the tensions in the string and the acceleration of the system.

R M T2 a m2 m2 g

T1 m1 a

m1 g

8.

A uniform disc of mass M and radius R is initially at rest. Its axis is fixed through O. A block of mass m is moving with speed v1 on a frictionless surface passes over the disc to the dotted position. On coming in contact with the disc, it slips on it. the slipping ceases before the block loses contact with the disc, due to the high friction. Now prove that v 2 =

9.

v1

v2

m

m O

R

v1 1 + ( M / 2m)

Find the angular momentum and rotational kinetic energy of earth about its own axis. Find the duration for which this amount of energy can supply 1 kilowatt power to each of 3.5×109 persons on earth ? Mass of earth = 6.0×1024., radius = 6.4×103km.

10. A uniform solid wheel of mass m and radius 15 cm is free to rotate without friction about a fixed horizontal axis passing through its centre. A particle of equal mass m strikes it at a pint P after falling vertically downward as shown in the figure, and sticks to the wheel.

Particle

37.5 cm 60o

P

Find the maximum angular speed of the wheel. find the force on the axis when the angular speed of the wheel reaches a maximum. 11. A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle of 53º with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder.

N1 N2

53o

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LEVEL – II 1.

A 10-m-long uniform ladder rests against a wall and makes an angle of 600 with the horizontal floor. If it just starts to slip from this position, find its instantaneous axis of rotation.

2.

In a playground there is a small merry-go-round of radius 2 m and mass 10 kg. The radius of gyration is 1.5 m. A boy of mass 30 kg runs at a speed of 5 ms-1 tangential to the rim of the merry-go-round when it is at rest and then jumps on. Neglect friction and find the angular velocity of the merry-go-round and the boy.

3.

A hoop of radius r rotating about its own axis with an angular velocity ω0 is gently placed with its plane vertical onto a rough horizontal surface. Find the velocity of the centre of the hoop when it ceases to slip. At the initial moment, the velocity of the centre of the hoop was zero.

4.

A uniform sphere of mass 2 kg and radius 10 cm is released from rest on an extremely rough inclined plane which makes an angle of 300 with the horizontal. Deduce (i) its angular acceleration (ii) linear acceleration along the plane and (iii) kinetic energy as it travels 2m along the plane.

5.

On the flat surface of a disc of radius a, a small circular hole of radius b is made with its centre at a distance c from the centre of the disc. If mass of the uncut disc is M, calculate the moment of inertia of the cut disc about the axis of the circular hole.

6.

A thin horizontal uniform rod AB of mass m and length A can rotate freely about a vertical axis passing through its end A. After some time, a constant force F is applied to the end B of the rod in its plane such that it is perpendicular to the initial orientation of the rod. Find the angular velocity of the rod as a function of its rotation angle φ measured relative to its initial position.

7.

A uniform circular disc of mass m, radius r and centre O is free to rotate about a smooth, horizontal axis which is tangential to the disc at a point A. The disc is held in a vertical plane with A directly below O and is then allowed to fall to its equilibrium position. Find the angular velocity of the disc when its plane is next vertical.

8.

A uniform circular disc of mass m and radius a is rotating with constant angular velocity ω in a horizontal m is placed gently onto the disc at its plane about a vertical axis through its centre A. An insect P of mass 2 rim. After the steady state is attained, the insect starts walking radially with respect to the disc with a aω at t = 0. Find the angular velocity of the system as a function of time t. speed 2

9.

ABC is a triangular framework of three uniform rods each of mass m and length 2A . It is free to rotate in its own plane about a smooth horizontal axis through A which is perpendicular to ABC. If it is released from rest when AB is horizontal and C is above AB, find the maximum velocity of C in the subsequent motion.

10. A cord is attached to a toy truck standing on a horizontal plane and passes over a fixed, frictionless, light pulley. A load of mass 1 m = k g is attached to the end of the string. Find the acceleration 2 of the truck given that the mass of each wheel is m' = 0.4 kg and the mass of the truck is M = 1.4 kg. The wheels are solid discs and they roll without slipping.

m

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LEVEL – III 1.

Show that if a symmetrical body of radius R rolls without slipping along a horizontal plane with velocity v c and reaches the plane inclined at an angle θ , it starts up the inclined plane with velocity     1 − cos θ ( )  where k is radius of gyration. It may be assumed that there is sufficient friction to v 'c = v c 1 −   k2  1 + 2    R   

prevent slipping and that the body doesn’t jump. 2.

A uniform catherine wheel is mounted in a vertical plane and is free to rotate about an axis passing through its centre and perpendicular to its plane. The wheel is made up of combustible material and burns at a constant rate of α kg/s and the reaction force exerted by the burning gases equals F tangentially at its circumference . If the wheel starts from rest, show that the angular speed acquired when half of its mass 4F 2 −1 . has burnt is αR

(

3.

4.

5.

6.

)

A uniform m solid hemisphere of radius r and mass M is pulled by means of a string so that it moves with a uniform velocity. If µ is the coefficient of friction between the hemisphere and surface, find the angle of inclination of the hemisphere. The centre of mass of a hemisphere is located at a distance 3r/8 from the centre.

θ

A plank of mass m 1 with a uniform solid sphere of mass m 2 placed on it rests on a force F is applied to the plank. With what acceleration will the plank and the centre of the sphere move provided there is no sliding between the plank and the sphere? A uniform cylinder of radius r is rotating about its axis at the angular velocity ω0 . It is now placed into a corner as shown in figure. The coefficient of friction between the wall and the cylinder as well as the ground and the cylinder is µ . Show that the number of turns n, the cylinder completes before it stops, are given by n =

ωo

m

A horizontal plank of mass m is lying on a smooth horizontal surface. A sphere of same mass and radius r is spinned to an angular frequency ω0 and gently placed on the plank with its axis horizontal with its axis horizontal as shown in the figure. If coefficient of friction between the plank and sphere is µ , find the distance moved by the plank till the sphere starts pure rolling on the plank. The plank is long enough.

F

m

m2

α m1 F

W ALL

ω20 r (1 + µ2 )

8πµg (1 + µ )

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7.

8.

9.

A sphere of radius r is projected up an inclined plane for which 1 µ =   tan θ with a velocity v 0 and initial angular velocity 7 ω0 (v 0 > r ω0 ) . Prove that friction acts downward at first and upwards afterwards. Further prove that the total time of rise is v 0 + 4 ω0 r . 18 g sin θ

V0

ω0

θ

A slender rod of mass m and length A is released from rest from vertical position as shown in the figure. Determine as a function of θ the normal forces, which is exerted on the rod by the ground as it falls downward, assuming that it does not slip.

θ

One end of a uniform rod of mass M and length L is supported by a frictionless hinge which can with stand a tension of 1.75 Mg. The rod is free to rotate in a vertical plane. To what maximum angle should the rod be rotated from the vertical position so that when left, the hinge does not break.

10. A rod of length l is constrained to move in a vertical plane containing its length between two smooth vertical planes. When it is released from the vertical position by pulling its bottom gently, find its angular acceleration and angular velocity as a function of the angle θ made by it with vertical

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SECTION–II SINGLE CHOICE QUESTIONS 1.

2.

3.

A stone of mass 0.1 kg is projected with a velocity of 20 2 m/s from the origin of an xy co-ordinate axis system at an angle of 45° with the horizontal. The angular momentum of the stone about the origin after 1 second is (take g = 10 m/s2): (a) 10 kˆ (kg m2 / s)

ˆ m 2 / s) (b) –10 k(kg

(c) –10 ˆi (kg m / s2 )

(d) –10 ˆi(kg m / s 2 )

For the toppling of the shown fig. regular hexagon. The coefficient of friction must be :

(a) > 0.21

(b) < 0.21

(c) = 0.21

(d) < 0.21

A ring of mass 0.3 kg & radius 0.1 m & a uniform solid cylinder of mass 0.4 kg and of the same radius are given the same kinetic energy and released simultaneously on a flat horizontal surface. It is observed that they begin to roll as soon as they are released towards a wall which is at the same distance from both the ring and the cylinder. The rolling friction in both cases is negligible. Then (a) the cylinder will reach the wall first

4.

5.

(c) both will reach the wall simultaneously (d) none will reach the wall. G G G If F be a force acting on a particle having the position vector r and τ be the torque of this force about the origin, then GG GG GG GG (a) r .τ = 0 and F .τ = 0 (b) r .τ = 0 and F .τ ≠ 0 GG GG GG GG (c) r .τ ≠ 0 and F .τ ≠ 0 (d) r .τ ≠ 0 and F .τ = 0 . A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with its axis fixed along the y-axis and free to rotate about its own axis. The platform is given a velocity in the x-direction given by x = A cos ωt . There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is (a)

1 MR A ω2 2

(c) 2 MRA ω2 6.

(b) the ring will reach the wall first

(b) MR A ω2 (d) mr ωA 2 cos 2 ωt .

A solid cylinder of diameter D is mounted on a frictionless horizontal axle. A string is wrapped around it and a heavy block is attached to the free end of the string. The block is allowed to fall freely. If the speed of the block just before striking the ground be v, then (a) v ∝ D (c) v ∝

1 D

(b) v ∝ D 2 (d) v is independent of D.

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7.

A uniform circular disc of radius R with a concentric circular hole of radius plane. The fraction of its total energy associated with its rotational motion is

8.

9.

(a)

1 4

(b)

(c)

1 2

(d) 1.

R rolls down an inclined 2

5 13

A ring starts from rest and acquires an angular speed of 10 rad/s in 2 s. The mass of the ring is 500 g and its radius is 20 cm. The torque on the ring is (a) 0.02 Nm

(b) 0.20 Nm

(c) 0.10 Nm

(d) 0.01 Nm.

A loop and a disc have the same mass and roll without slipping with the same linear velocity v. If the total kinetic energy of the loop is 8 J, the kinetic energy of the disc must be (a) 6 J

(b) 8 J

(c) 10 J

(d) 12 J.

10. A wheel and an axle, having a total moment of inertia 0.002 kg-m2, is made to rotate about a horizontal axis by means of an 800 g mass attached to a cord (assumed massless) that is wound around its axle. The radius of the axle is 2 cm. Starting from rest, how far does the mass fall in order to give the wheel a speed of 3 rev/s? (a) 2.25 cm

(b) 3.25 cm

(c) 4.5 cm

(d) 5. 25 cm.

11. Two identical solid uniform spheres A and B of mass m and radius r each, are placed with their centres on the x-axis separated by a distance 3r. Sphere A is spun with an angular velocity ω about an axis parallel to the y-axis passing through its centre. Sphere B is spun similarly with an angular velocity ω , but about the z-axis. The total angular momentum of the system about the x-axis is (a)

2 2 mr 2 ω 5

(c) zero

(b) (d)

4 mr 2 ω 5 53 mr 2 ω . 5

12. A small meteorite of mass m travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass M and radius R. The length of the day was T before the meteorite struck. After the meteorite strikes the earth, the length of day increases (in sec) by (a)

5m T 2M

(b)

m T M

(c)

4m T 5M

(d)

M . 3m T

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13. A uniform rod of length A is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle θ , its angular velocity is given by (a)

θ  6g    sin 2  A 

(b)

 6g   A

(c)

 6g   A

(d)

 6g    cos θ .  A 

  sin θ 

θ   cos 2 

A

l

A|

θ B

14. A disc of mass M and radius R is pivoted about a horizontal axis through its centre and a particle of the same mass M is attached to the rim of the disc. If the disc is released from rest with the small body at the end of a horizontal radius, the angular speed when the small body is at the bottom is: (a)

 g     4R 

(b)

 g     2R 

(c)

 3g     4R 

(d)

 4g   .  3R 

15. A symmetric lamina of mass M consists of a square shape with a semicircular section over each of the edge of the square as shown in figure. The side of the square is 2a. The moment of inertia of the lamina about an axis through its centre of mass and perpendicular to the plane is 1.6 M a 2 . The moment of inertia of the lamina about the tangent AB in the plane of the lamina is: (a)

4.8 Ma 2

A 2a

3.2 Ma 2

(b)

B (c) 6.4 Ma 2 (d) 1.6 Ma 2 . 16. A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v . It collides elastically and head-on with an identical sphere B which is at rest. All surfaces are frictionless. After the collision, their angular speeds are ωA and ωB , respectively. Then,

(a) ωB = 0

(b) ωA = ωB

(c) ωA < ωB

(d) ωB = ω .

17. A cubical block of side a is moving with a velocity v on a smooth horizontal plane as shown in the figure. It hits a ridge at point O. The angular speed of the block after it hits O is: (a) (c)

3v 4a 3v 2a

(b)

3v 2a

O

(d) zero.

18. A solid uniform sphere, rotating about a horizontal axis (with rotational K.E. = E 0 ), is gently placed on a rough horizontal plane. After some time the sphere begins pure rolling with total Kinetic Energy E. Then, 52 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016

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(a) E =

2E 0 5

(b) E =

2E 0 7

(c) E =

5E 0 7

(d) none of the above is true.

19. The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is not given by : (b) I 3 + I 4 (a) I 1 + I 2 (c) I 1 + I 3 (d) I 1 + I 2 + I 3 + I 4 where I1, I2, I3, I4 are the moments of inertia of the plates about axes 1,2,3,4 respectively.

A

B O

1 D

3

C

2

4

20. A circular table has a radius of R and mass M. It has 4 legs of length L each fixed symmetrically on its circumference. The maximum mass m (if possible) which can be placed anywhere on this table without toppling it is

( M (

(a) M (c)

) 2 + 1) 2 −1

(b) 4 M (d) cannot be determined.

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SECTION–III MULTIPLE CHOICE QUESTIONS 1.

Which of the following statement(s) is/are correct for a spherical body rolling without slipping on a rough horizontal ground at rest. (a) The acceleration of a point in contact with ground is zero (b) The speed of some of the point(s) is (are) zero (c) Friction force may or may not be zero (d) Work done by friction may or may not be zero

2.

A uniform rod of length 6a and mass 8m lies on a smooth horizontal table. Two particles of mass m and 2m, moving in the same horizontal plane but in opposite directions with speeds 2v and v respectively strke the rod normally as shown in figure and stick to the rod. Denoting angular velocity (about the centre of mass), total energy and translational velocity of centre of mass by ω, E and vc respectively after the collision. (b) ω =

(a) vc = 0 (c) ω = 3.

v

(d) E =

5a

m

2v a

2a

2m v

3v 5a 3mv 2 5 →

→

A particle moves in acircle of radius r with angular velocity ω ω. At some instant its velocity is v and →

→

radius vector with respect to centre of the circle is r . At this particular instant centripetal acceleration ac of the particle would be →

→

→

→

(a) ω × v

→



(c) ω ×  ω × r    4.

→

→

→

→

(d) v × (r × ω ) →

A particle of mass m is travelling with a constant velocity v = v 0iˆ along the line y = b, z = 0. Let dA be the area swept out by the position vector from origin to the particle in time dt and L the magnituded angular momentum of particle about origin at any time t. Then (a) L = constant (c)

5.

→

(b) v × ω

dA

=

2L

(b) L ≠ constant (d)

dA

L

=

2m dt dt m In pure rolling fraction of its total energy associated with rotation is a for α ring and β for a solid sphere. Then (a) α = (c) β =

1 2

2 5

(b) α = (d) β =

1 4

2 7

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6.

7.

A disc can roll without slipping without applying any external force on a (a) rough inclined plane

(b) smooth inclined plane

(c) rough horizontal surface

(d) smooth horizontal surface

A disc of radius R rolls on a horizontal surface with linear velocity v and angular velocity ω. There is a point P on the circumference of the disc at angle θ which has a vertical velocity. Here θ is equal to

 v   Rω 

(b)

 v   Rω 

(d) π + cos-1  

-1 (a) π + sin  

(c) π − sin -1   8.

π 2

 v   Rω 

− sin -1  

P θ O

ω v

 v   Rω 

A special body of radius R rolls on a horizontal surface with linear velocity v. Let L1 and L2 be the magnitudes of angular momenta of the body about centre of mass and point of contact P. Then (a) L2 = 2L1 if radius of gyration K = R

v

ω

(b) L2 = 2L1 for all cases

P

(c) L2 > 2L1 if radius of gyration K < R (d) L2 > 2L1 if radius of gyration K > R 9.

The axis of rotation of a purely rotating body (a) must pass through the centre of mass (b) may pass through the centre of mass (c) must pass through a particle of the body (d) may pass through a particle of the body

10. A particle moves on a straight line with a uniform velocity. Its angular momentum (a) is always zero (b) is zero about a point on the straight line (c) is not zero about a point away from the straight line (d) about any given point remains constant.

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SECTION–IV PASSAGE BASED QUESTIONS # Write-up I [Questions 1 to 3] A disc of mass M and Radius R is given a velocity Vo to its centre and an angular velocity of ω0 about centre as shown. Then it is kept over a rough surface where coefficient of friction is µ. Initially the lower most point or point of contact slides over the surface (on which it is kept) and friction acts. Due to friction velocity of centre of mass of disc is affected as well as it will produce a torque which changes the angular velocity of body about centre. After some time rolling is established. 1.

2.

If

ω0 R = v0

ω0

then time after which rolling starts is

3v0 µg

(a)

v0 µg

(b)

(c)

2 v0 3 µg

(d) rolling will not start

Velocity of centre of disc at the moment when angular velocity of disc becomes zero is (a) Vo (c)

3.

positive x-direction V0

R

V0 2

(b)

V0 3

(d) 2V0

When rolling starts then (a) friction will not act

(b) friction will act along positive direction

(c) friction will act along negative direction

(d) we can’t say about friction

# Write-up II [Questions 4 to 6] A small particle of mass m is given an initial velocity v0 tangent to the horizontal rim of a smooth cone at a radius r0 from the vertical centerline as shown at point A. As the particle slides to point B, a vertical distance h below A and a distance r from the vertical centerline, its velocity v makes an angle θ with the horizontal tangent to the cone through B.

4.

A

v0 r0

θ

r B

v α α

The value of θ is (a) cos

−1

(c) cos

−1

v 0 r0 v02 + 2gh (r0 − h tan α ) v0 r0 v02 − 2gh (r0 − h tan α)

(b) cos

−1

(d) cos

−1

v0 r0 v02 + 2gh (r0 + h tan α ) v0 r0 r0 v02 + 2gh

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5.

6.

The speed of particle at point B (a)

v 20 + 2gh

(b)

v 20 − 2gh

(c)

v 20 + gh

(d)

2 v02 + 2gh

The minimum value of v0 for which particle will be moving in a horizontal circle of radius r0. (a)

2gr0 tan α

(b)

gr0 2 tan α

(c)

gr0 tan α

(d)

4gr0 tan α

MATCHING TYPE QUESTIONS 7.

A rod of length L and weight w is kept in equilibrium on the L as shown in the figure. The right two support separated by 2 support is taken out at time t = 0.

w, L L 2

Match the following questions based on the above information Column I (A)

(B)

Column II

The moment of inertia of the rod about the support point at t = 0 is

(P)

The angular acceleration of rod about the

(Q)

3g 7 12 g 7L

support point at t = 0 is (C)

The linear acceleration of centre of mass of rod at t = 0 is

(R)

4 w 7

(D)

The normal reaction on the rod by the support at t = 0 is

(S)

7 wL2 48 g

(T)

wL2 3g

(U)

3g 7L

(V)

w 4

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8.

For the following statements, except gravity and contact force between the contact surfaces, no other force is acting on the body. Column I

Column II

(A)

When a sphere is in pure–rolling on a fixed horizontal surface.

(P)

Upward direction

(B)

When a cylinder is in pure rolling on a fixed inclined plane in upward direction then friction force acts in

(Q)

vcm > R ω

(C)

When a cylinder is in pure rolling down a fixed incline plane, friction force acts in

(R)

vcm < Rω

(D)

When a sphere of radius R is rolling with slipping on a fixed horizontal surface, the relation between vcm and ω is

(S)

No frictional force acts.

ASSERTION-REASON TYPE QUESTIONS Following questions consist of two statements printed as Assertion and Reason. While answering these questions you are required to select any one of the responses indicated as (A) (B) (C) (D)

If both assertion and reason are true and reason is a correct explanation of the assertion. If both assertion and reason are true but the reason is not a correct explanation of assertion. If assertion is true but reason is false. If assertion is false but reason is true.

9.

Assertion : Moment of inertia of circular ring about a given axis is more than moment of inertia of the circular disc of same mass and same size, about the same axis. Reason : The circular ring is hollow so its moment of inertia is more than circular disc which is solid. (a) A

(b) B

(c) C

(d) D

10. Assertion : The total kinetic energy of a rolling solid sphere is the sum of translational and rotational kinetic energies Reason : For all solid bodies total kinetic energy is always twice the translational kinetic energy (a) A

(b) B

(c) C

(d) D

11. Assertion : Two circular discs of equal masses and thickness made of different material, will have same moment of inertia about their central axes of rotation. Reason : Moment of inertia depends upon the distribution of mass in the body. (a) A

(b) B

(c) C

(d) D

12. Assertion : A ladder is more apt to slip, when you are high up on it than when you just begin to climb. Reason : At the high up on a ladder, the torque is large and on climbing up the torque is small. (a) A

(b) B

(c) C

(d) D

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13. Assertion : When a sphere and a solid cylinder are allowed to roll down an inclined plane, the sphere will reach the ground first even if the mass and radius of the two bodies are different. Reason : The acceleration of the body rolling down the inclined plane is directly proportional to the radius of the rolling body (a) A

(b) B

(c) C

(d) D

14. Assertion : A hollow shaft is found to be stronger than a solid shaft made of same material Reason : The torque required to produce a given twist in hollow cylinder is greater than that required to twist a solid cylinder of same size and material (a) A

(b) B

(c) C

(d) D

15. Assertion : Angular velocity is a characteristic of the rigid body as a whole Reason : Angular velocity may be different for different particles of a rigid body about the axis of rotation. (a) A

(b) B

(c) C

(d) D

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SECTION–V PROBLEMS ASKED IN IIT-JEE SUBJECTIVE 1.

A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R/2. [IIT –1990]

2.

A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed at a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R/4 from the center C of the disc. The line AC is perpendicular to PQ. Initially, the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position. [IIT –1998]

3.

A R C

R/4 Q

P

Two thin circular discs of mass 2 kg and radius 10 cm each are joined by a rigid massless rod of length 20 cm. The axis of the rod is along the perpendicular to the planes of the disc through their centers. This object is kept on a truck in such a way that the axis of the object is horizontal and perpendicular to the direction of the motion of the truck. Its friction with the floor of the truck is large enough so that the object can roll on the truck without slipping. Take x–axis in the direction of motion of the truck and z-axis in the vertically upward direction. If the truck has an acceleration of 9 m/s2,calculate

O

(i) The force of friction on each disc. (ii) The magnitude and the direction of the frictional torque acting on each disc about the center of mass O G G G of the object. Express the torque in the vector form in terms of unit vectors i , j and k in the x, y and z directions. [IIT –1997] 4. A block X of mass 0.5 kg is held by a long massless string on a frictionless inclined plane of inclination 30° to the horizontal. The string is wound on a T uniform solid cylindrical drum Y of mass 2 kg and radius 0.2 m as shown in T figure. The drum is given an initial angular velocity such that the block X starts X moving up the plane. (a) Find the tension in the string during the motion. (b) At a certain instant of time, the magnitude of the angular velocity of Y is 10 rad s-1. Calculate the distance traveled by X from that instant of time until it 30o comes to rest. [IIT –1974] 5.

A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v respectively, strike the bar (as shown in the figure) and stick to the bar after collision. Calculate (a) velocity of the center of mass (b) angular velocity about center of mass and (c) total kinetic energy, just after collision. [IIT –1991]

2m v

C 8m

2a

a

2a

2v

a

m

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6.

A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is 0.16 kg and length is

( 3)

m. Two particles,

each of mass 0.08 kg are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with a velocity of 10 m/s and the other with 6 m/s as shown in the figure. The first particle strikes the bar at point A and the other at point B. Point A and B are at a distance of 0.5 m from the center of the bar. The particles strike the bar at the same instant of time and stick to the bar on collision. Calculate the loss of KE of the system in the above collision process.

A

10m/s

6m/s

B

[IIT- 1989] 7.

A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius R is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane. There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine :

R

(a) The angle θc through which the cylinder rotates before it leaves contact with the edge. (b) The speed of the center of mass of the cylinder before leaving contact with the edge and (c) The ratio of translational to rotational kinetic energies of the cylinder when its center of mass is in horizontal line with the edge. [IIT – 1995] 8.

A homogeneous rod AB of length L = 1.8m and mass M is pivoted at the center O in such a way that it can rotate freely in the vertical plane as shown in the figure. The rod is initially in the horizontal position. An insect S of the same mass M falls vertically with speed V on the point C, midway between the points O and B. Immediately after falling, the insect moves

S L 4

A

L 2

C

L 4

B

towards the end B such that the rod rotates with constant angular velocity ω . (a) Determine the velocity ω in terms of V and L. (b) If the insect reaches the end B when the rod has turned through an angle of 90°, determine V. [IIT-1992] 9.

A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m traveling along the surface hits the end A of the rod with a velocity v0 in a direction perpendicular to AB. The collision is completely elastic. After the collision the particle comes to rest : (a) Find the ratio m/M. (b) A point P on the rod is at rest immediately after the collision. Find the distance AP. [IIT-2000] (c) Find the linear speed of the point P at time πL ( 3v0 ) after the collision. 10. Two heavy metallic plates are joined together at 90° to each other. A laminar sheet of mass 30 kg is hinged at the line AB joining the two A heavy metallic plates. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel to AB and passing Q through its center of mass is 1.2 kg-m2. Two rubber obstacles P and Q B are fixed, one on each metallic plate, at a distance 0.5 m from the line P AB. This distance is chosen so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and turns back. If the impulse on the sheet due to each obstacle is 6 N-s (a) Find the location of the center of mass of the laminar sheet from AB. (b) At what angular velocity does the laminar sheet come back after the first impact ? (c) After how many impacts does the laminar sheet come to rest ? [IIT-2001] 61 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016

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OBJECTIVE 1.

a=0.6m

A plate of length a = 0.6 m and width b = 0.4 m having a mass M = 3 kg is hinged at one of its edges AB as shown in the figure. The plate is held horizontal due to small particles colliding elastically with it from below. The particles, each of mass m = 0.01 kg strike the shaded half of the plate elastically at the rate of 100 particles per unit area per unit time. The velocity of the striking particles is [IIT-2006]

2.

(a) 83.33 m/s

(b) 80 m/s

(c) 75.5 m/s

(d) 70 m/s

b=0.4m

A solid sphere of radius R is moulded into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis perpendicular to the cross section of the disc and passing through circumference is equal to the moment of inertia of the sphere about its diameter. The value of r (in terms of R) is [IIT 2006]

3.

(a)

2 R 15

(b)

2 R 15

(c)

1 R 2

(d)

1 [ ]R 5

A small ball starts rolling from position A down the fixed hemispherical surface (see figure) to position C. There is friction on path AB while path BC is frictionless. If KA, KB, KC represent kinetic energy of the ball and [IIT 2006] hA, hB, hC height of the ball at points A, B and C, then (a) hA = hC , K A > K C

(b) hA < hC , K C > K A

(c) hA > hC , K A < K C

(d) hA > hC , K C > K A

C hC

hA A B

PASSAGE BASED QUESTIONS (From Q. 4 to Q.6) Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential energy of a spring compressed by a distance x1. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x2. Both the discs rotate in the clockwise direction. [IIT 2007] 4. The radio x1/x2 is (a) 2 (c) 5.

(b) 2

(d)

1 2 1 2

.

When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is (a)

2I ω 3t

(b)

9I ω 2t

(c)

9I ω 4t

(d)

3I ω 2t

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Physics : Rotational Motion

INSTITUTE OF CORRESPONDENCE COURSES

6.

7.

8.

The loss of kinetic energy during the above process is (a)

I ω2 2

(b)

I ω2 3

(c)

I ω2 4

(d)

I ω2 . 6

A small object of uniform density rolls up a curved surface with an 3v 2 with initial velocity v. It reaches up to a maximum height of 4g respect to the initial position. The object is (a) ring

[IIT 2007] (b) solid sphere

(c) hollow sphere

(d) disc

v

STATEMENT – 1 If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant. BECAUSE STATEMENT–2 The linear momentum of an isolated system remains constant [IIT 2007] (a) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1 (b) Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct explanation for Statement–1 (c) Statement–1 is True, Statement–2 is False (d) Statement–1 is False, Statement–2 is True

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Physics : Rotational Motion

INSTITUTE OF CORRESPONDENCE COURSES

ANSWERS SECTION - I SUBJECTIVE TYPE QUESTIONS LEVEL – I 1.

G aB − bA τ = ( aB − bA) kˆ , A ⊥ = A 2 + B2

2.

V = 5.4 m / s

3.

θ = 6 rad , ω = 6 rad / s ,

L = 7.1 × 1023 kg m 2 / s

9.

K = 2.6 × 1029 joule , t = 2.35 × 109 year 10. ω max =

v = 3 m / s , a t = 1.5 m / s 2 , a R = 18 m / s 2 4.

 mm  I =  1 2  r 2 ; I = 2.79 × 10−47 kg-m2  m1 + m2 

6.

a = 1.57m / s 2 , T1 = 99N,T2 = 91 N

7.

v2 =

Fmax =

20 5 rad / s 3

16mg 3

11. N 2 = W = 100N; f=N1 =

200 N 3

v1 M   m + 2m  LEVEL – II

1.

5m and 5 3 m from vertical & horizontal walls respectively. 2. ω = 2.1 rad/s 3 . v = 0.5 ω0 r 4. (i) 35 rad/s2 , (ii) 3.5 m/s2, (iii) 10 J  b4  1 M a 2 + 2c 2 − 2  5. 2 a   6.

g 5r

7.

ω= 4

8.

ω′ =

9.

2 gA 3

10.

mg = 1.15 m/s2 M + 6m ′ + m

ω 2

 ω  1 + 1 − t   2 

6F sin φ mA LEVEL - III

3. 4. 5.

 8µ  θ = sin −1    3 + 8µ  2ω20 r 2 81µg F F , 2 7 m 1 + m 2 m 2 + m1 7 2

N =

8.

mg 9 cos 2 θ − 6 cos θ + 1 4

(

 3cos θ − 1  = mg   2  

)

2

θ = 60º 3 g (1 − cos θ) 10. w = A 9.

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Physics : Rotational Motion

INSTITUTE OF CORRESPONDENCE COURSES

SECTION – II SINGLE CHOICE QUESTIONS 1.

(a)

2.

(a)

3.

(c)

4.

(a)

5.

(a)

6.

(d)

7.

(b)

8.

(c)

9.

(a)

10.

(c)

11.

(c)

12.

(a)

13.

(a)

14.

(d)

15.

(a)

16.

(a)

17.

(a)

18.

(b)

19.

(d)

20.

(d)

SECTION – III MULTIPLE CHOICE QUESTIONS 1.

(b,c)

2.

(c,d)

3.

(a,c)

4.

(a,d)

5.

(d)

6.

(a,c,d)

7.

(c,d)

8.

(a,d)

9.

(b,d)

10.

(d)

SECTION - IV PASSAGE BASED QUESTIONS 1.

2.

(a)

(b)

3.

4.

(a)

(a)

5.

6.

(a)

(c)

MATCHING TYPE QUESTIONS 7.

8.

A→S B→Q C→P D→R

A→S B→P C→P D→Q,R

ASSERTION-REASON TYPE QUESTION 9.

(b)

10.

(c)

11.

(d)

13.

(c)

14.

(a)

15.

(c)

12.

(a)

SECTION –V PROBLEMS BASED IN IIT-JEE SUBJECTIVE 1. 2. 3

(14 g R /3) 5gR G (i) f = (6iˆ)N G G (ii) τ1 = 0.6( kˆ − j ) & τ2 = 0.6( − jˆ − kˆ ) 65 FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016

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Physics : Rotational Motion

INSTITUTE OF CORRESPONDENCE COURSES

4. 5

6. 7.

(a) (b) (a) (b)

1.63 N 1.22 m zero ω = (v /5a )

3 (c) K.E. = mv 2 5 2.72 J 4 (a) θc = cos−1   7 (b)

4 gR 7

(c)

KT =6 KR

 12  V (a) ω =   7 L 7 (b) V = 2 gL = 3.5 m/s 12 1 9. (a) 4 2 (b) = L 3 v G (c) |v P |= 0 2 2 10. (a) 0.1 m (b) 1 rad/s (c) Infinite . 8.

1. 5.

(a) (a)

OBJECTIVE 2. 6.

(a) (b)

3. 7.

(c) (d)

4. 8.

(c) (d)

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