Rotational Motion-th

1.

IITJEE Syllabus

Moment of inertia, parallel and perpendicular axes theorems, moment of inertia of bodies having simple geometrical shapes, e.g. uniform ring, disc, thin rod and cylinder. Angular momentum and torque, conservation of angular momentum. Equilibrium of rigid bodies

2.

Rotational Kinematics

A rigid body is a body with a definite and unchanged shape and size i.e. a body is said to be rigid if the distance between any two particles of the body remains invariant.

2.1

Motion of a rigid body

Translation: A rigid body is said to undergo translation if it moves such that it always remains parallel to itself: this means that a line connecting any two particles of the rigid body always remains parallel to itself throughout its motion. Rotation: A rigid body is said to undergo rotation if there exists a straight line from which the distance of any particle of the rigid body remains constant throughout its motion. This straight line, whether fixed or moving is known as the axis of rotation. The rigid body is said to undergo rotation about this axis. Angular Displacement: Consider a rigid body undergoing rotation about an axis, perpendicular to the plane of the paper and passing through O. Suppose that A and B are any two particles of the rigid body at the position 1 while A and B are their subsequent locations when the body is at the position 2. Since the body undergoes rotation, OA = OA and OB = OB Further AB = AB, since the body is rigid.  OAB OAB(congruent) i.e. AOB = AOB Adding AOB to both sides of the above equation, we get, BOB = AOA =  (say) This implies that in a given interval of time the angular displacements of all particles of the rigid body undergoing rotation are identical. Therefore, a single variable, viz. angular displacement () can be used to describe the rotational motion of the rigid body.

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Angular displacement is not a vector quantity. However, for infinitesimal time intervals, the corresponding angular displacement is infinitesimal and behaves like a vector. Average angular velocity is defined for a fixed axis of rotation, by

avg 

 2  1 , t 2  t1

Instantaneous angular velocity is defined by  

d , the direction of  (instantaneous) dt

ˆ. is along the axis n

Average and instantaneous angular accelerations are defined by     2  1  d  avg    , t 2  t1 dt Angular velocity is same for all the particles of a rigid body and the same is true about angular acceleration also as and the reason being equal angular displace accment. If the angular acceleration is constant, the following relations hold: (t) = o + t (t) = o + ot +

1 2 t 2

  t  =  20  2 2

Here o = magnitude of the initial angular velocity (t) = magnitude of the angular velocity after time t. o = Initial angular position. (t) = Angular position after time t. Illustration 1:

A fan starts rotating with constant angular acceleration of  radians/s2 about a fixed axis perpendicular to its plane and through its centre. Find (a) the angular velocity of the fan after 4 sec. (b) the angular displacement of the fan after 4 sec and (c) number of turns accomplished by the fan in 4 sec.

Solution :

Here

=  rad/sec2 0 = 0 t = 4 sec (a) (4 sec) = 0 + ( rad/sec2)  4 sec = 4 rad/sec. (b) (4 sec) = 0 +

1 ( rad/sec2)  (16 sec2) = 8 radian. 2

(c) Let the number of turns be n  n 2 rad = 8 rad



n=4

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A wheel rotates with an angular acceleration given by  = 4at3 - 3bt2 , where t is the time and a and b are constants. If the wheel has initial angular speed 0, write the equations for the: (i) angular speed (ii) angle displacement.

Exercise 1:

2.2

Relation between linear and angular variables

Consider a particle A of a rigid body undergoing rotation ˆ , the particle A describing an arc about a fixed axis- n ABA of a circle with its centre O on the axis of rotation. Taking the origin at O, the position vector of A,   OA . r OA = OA = constant (radius of the circle) AOA = (t) (say) The arc length, ABA, S = r dS d  r = r The tangential velocity, vA = dt dt  The direction of the angular velocity vector  be taken along the axis of rotation:  ˆ, n ˆ being the unit vector along the axis of rotation.  = n  Then, v , instantaneous velocity of A with respect to the axis of rotation, can be written as    v   r

The acceleration of the point A with respect to the axis of rotation is 





 dv d   d r   r    a=

If



dt dt dt    is constant, then  = 0 and,        a    v =      r  =  2 r

     a    r   v

Illustration 2:

A disc rotates with constant angular acceleration of 2 rad/s 2 about a fixed vertical axis through its centre and perpendicular to its plane. A coin is placed on it at a distance of 1m from the axis of rotation. The coin is always at rest relative to the disc. If at t = 0 the disc was at rest, then find the total acceleration of the coin after one second.

Solution:

After 1 second angular velocity of the disc and hence that of the coin about the axis of rotation is = 0 + 2(rad/s2)  1 sec = 2 rad/s aT = r = (2 rad/s2)  1 m = 2m/s2. aR = 2 r = (2rad/s)2  1m = 4 m/s2 

a=

2 a2T  aR  2 5m / s2

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3.

Rotational dynamics

3.1

Kinetic energy of rotation and moment of inertia :

A rigid body undergoes pure rotational motion about a fixed axis n and its constituent particles move on circular paths r1 v1 with radii r1, r2, ... and rn (say) with linear velocities v1 = r1, r2 v2 = r2, ..... and vn = rn,  being the angular velocity of the v2 r3 rigid body. v3 If m1, m2, ......... and mn are the masses of the respective particles of the rigid body, then the kinetic energy of the system is given by 1 1 1 m1v12  m 2 v 22  ..........  mn v n2 K.E. = 2 2 2 1 1 1 m12r12  m22r22  .......... .. = m1r12  m2r22  ..........  mnrn2 2 = 2 2 2



The term

m1r12  m 2r22  .......... ....  mnrn2



is called rotational inertia or moment of

inertia of the body or system of particles. Rotational inertia of a particle of mass ‘m’ is given by the following expression I = mr2 ; where r = perpendicular distance of the particle from the axis of rotation. Rotational inertia of a system of particles is given by the expression n

I=

 miri2 i 1

For a continuous distribution, I =  r 2dm where dm is a small element of the body at a distance r, from the axis of rotation. The kinetic energy of a rigid body, due to rotation, is given by 1 K.E. = I2 2 Moment of inertia of a system of particles depends on : (i) Axis of rotation (ii)

Mass of the system

(iii)

Distribution of mass in the body

The moment of inertia of a rigid body about a given axis of rotation, is a constant. Moment of inertia plays the same role in rotational motion as mass plays in translational motion. Illustration 3:

Calculate the moment of inertia of a thin ring of mass 'm' and radius 'R' about an axis passing through its centre and perpendicular to the plane of the ring .

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Solution :

Remember that in case of continuous mass distribution, we use the 2 formula I =   dm  r to find out the moment of inertia of the body AA is the axis about which rotation of the ring is being considered. Mass of the ring = M, length of the ring 2R, Mass per unit length of 

M 

 = (say) the ring =   2 R  A

Consider a small element of the ring at a polar angle  from a particular reference radius. The element subtends an angle d at the centre. Length of the element = Rd Mass of the element ( Rd) Moment of inertia of the element=( Rd) R2 Moment of inertia of the ring =



2

0

M

R d

I = mR2

  Rd  R 2

A

= MR2 Exercise 2:

Calculate the moment of inertia of a disc of radius R and mass M, about an axis passing through its centre and perpendicular to the plane

Moment of Inertia of some Common Bodies (a) Ring

(b) Disc

(c) Cylinder m

m

R

m

R

R

L

I = mR2

mR 2 I= 2 I=

(d) Rod

(e) Rod

mR 2 2

(f) Solid Box a L

L I=

c I=

mL2 12

(g) Solid Sphere

mL2 3

b I=

m(a2+b2) 12

(h) Hollow sphere

R m

I

2 mR 2 5

I

2 mR 2 3

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Radius of Gyration: The radius of gyration of a body is that distance from the axis of rotation, where a particle of equal mass should be placed so as to possess the same moment of inertia as the given mass distribution. Just as the centre of mass can be used to represent the translational motion of a rigid body, its radius of gyration can be used to study rotation. I m

For a body of mass m and moment of inertia I, radius of gyration k =

3.2

Parallel Axis Theorem

If the moment of inertia of a rigid body about an axis(n) passing through its centre of mass is Icm, then the moment of inertia of the body about an axis parallel to n, at a distance d from the first one, is given by I = Ic.m. + md2 This theorem is known as the parallel axis theorem. Illustration 4:

m G d

Icm

Using the parallel axis theorem, find the M.I. of a sphere of mass m 2 mr 2 . about an axis that touches it. Given that I cm  5

Solution : 

Ip = Icm + m (OP)2 Ip = I0 + m (OP)2 =

3.3

1

2

2 7 m r2 + mr2 = mr2 5 5

Perpendicular Axis Theorem

If the moment of inertia of a plane lamina about two mutually perpendicular axes in its plane are Ix and Iy, then its moment of inertia about a third axis (z) perpendicular to both the axes and passing through the point of intersection is IZ  IX  IY This theorem is known as the perpendicular axis theorem. Illustration 5 :

Using perpendicular axes theorem, find the M.I. of a disc about an axis passing through its diameter.

Solution :

According to perpendicular axis theorem, IZ

I X I Y

we know that I X  I Y due to the geometrical symmetry of the disc.



IZ  IX  IY  IX  IY 

IZ 2

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where I Z  M.I. of the disc about z axis passing through its center 1 perpendicular to its plane  mr 2 2 1 M.I.  I X  I Y  m r 2  4

3.4

Torque

Torque is the turning or twisting action of a force, acting on a body, about its axis of rotation.   If a force F is acting at a point P on a rigid body, then the torque due to the force F acting on the body, measured about the axis of rotation (n), is given by      r  F  , where r is p.v. of pt. P w.r.t. origin O. Magnitude of the torque is equal to the product of the force and the shortest distance between the force and the axis of rotation.    = rF sin where  = angle between r and  F.

Illustration 6:

A force N

is

 r 

 ˆ ) F  (2ˆ i  3ˆ j  4k

applied

to

a

point

having

ˆ) ( 3ˆ i  2ˆ j k

position vector Find the

m .

torque due to the force about the origin.

Solution:

     r  F

ˆ )  ( 2ˆ  (3 ˆ i  2ˆ j k i 3 ˆ j 4

) ˆ i  3 2

ˆj 2 3

ˆ k 1 4

ˆ ( 9  4)  [ ˆ i ( 8  3 )  j(12 2 )  k  ˆ ) N-m   (5 ˆi  10 ˆj  5k

  ( 5)2 ( 10)2 (5)2 N  m 

Exercise 3:

 = 56 N – m.

A particle describes uniform circular motion. Find the torque acting on it about the centre of the circle.

Work done by a Torque

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Work done by torque , W =  ;  = angular displacement For variable torque, dW =  d and W = d Illustration 7 :

Solution:

A tangential force of F = 1.5 N acts on a particle of mass m = 2 kg revolving in a circular path of radius r = 3m. What is the work done by the torque for a complete revolution of the particle?

Work done by the torque  is given by W   d where   rF



W  rF



 d 0

Putting   2  for a complete revolution we obtain W  2rF



W  2 ( 3 )(1.5 )9  Exercise 4:

J

Referring to the previous illustration, what is the work done by the torque during the first second?

Pseudo Torque : If motion of the body is observed from a non-inertial frame of reference having an acceleration a in a fixed direction with respect to an inertial frame, one has to apply a pseudo force (-ma) to each particle. These pseudo forces produce a pseudo torque about the axis. In such a case ext = I does not hold. But there exists a very special and very useful case when ext = I does hold even if the angular acceleration  is measured from a non-inertial frame A. That special case occurs when the axis of rotation passes through the centre of mass.

3.5

Angular momentum

For a particle having momentum 

 p,

the angular momentum about the point O is p  Here r = position vector of the particle with respect to the given point   L  r p



and p = linear momentum of the particle. r  The direction of L can be determined by using the rule for vector O products. The magnitude of the angular momentum of a particle about a fixed point is equal to the product of its linear momentum and the length of the perpendicular to the linear momentum from the fixed point O. Exercise 5:

Find the angular momentum of a particle of mass m describing a circle of radius r with angular speed .

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Illustration 8 :

A particle of mass m is projected with a velocity v at an angle with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory.

Solution :

At the highest point it has only horizontal velocity vx = v cos

y

Length of the perpendicular to the horizontal velocity from ‘O’ is the maximum height, where Hmax = 

H

x

O

v 2 sin2  2g

Angular momentum L=

mv 3 sin2  cos  2g

Angular momentum of a rigid body: Suppose that a rigid body describes pure rotational motion. All its constituent particles also describe circular motion about the same axis of rotation. The angular momentum of the particles about their corresponding centres are given by 2 2 L1 = m1r1 , L 2  m2r2

r1 r2

and similarly Ln = mn rn2 where is the angular

r3

v1 v2 v3

speed of the body. Therefore angular momentum of the whole body is given by L = L1 + L2 + ........... + L n =

m r

2 11



 m 2r22  ..........  mnrn2  , as the angular momentum of all the

particles are in the same direction i.e. along the axis of rotation.  L = I Relation between torque and angular momentum 

   L  r p

       dL d r   dp  p  r   v   mv   r  F dt dt dt    = 0 + r  F  ext   dL   ext dt



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  dp  Fext This relation is analogous to dt

3.6

Conservation of Angular Momentum

When there is no net external torque acting on a particle, then



 L = constant

 dL  0. dt

Therefore, the angular momentum of the particle remains invariant in the absence of any net external torque. Condition under which Torque acting on a body is Zero   r Fsin   0 , when r = 0, that means that the force passes through the axis of

rotation.   0 , when F = 0 (or negligibly small) that means no external force acts on

the particle.  0

when  = 00 or 1800 the force is parallel to the radius vector. (Radial forces cannot impart any torque on the particle i.e. centripetal force causes no torque). When any or all of the above conditions are satisfied,  = 0 and thus the angular momentum of the particle is conserved. That means the particle will keep on rotating with a uniform or constant angular velocity. Exercise 6 :

What is the moment of the gravitational force of the sun on earth about the axis of its rotation about the sun?

Relation between torque and angular acceleration : Suppose that the angular speed of a body changes from to (+) in a very small time interval t. If the angular displacement of the body during this time interval is =(t) then



from the Work Energy Theorem. 1  = I(2 ), 2 where  = average torque during this time interval.  = K.E. (of rotation)  t = I (as = t)   =I  t Similarly instantaneous torque is given by

 =I

 = I Illustration 9:

A uniform disc of radius R and mass M is free to rotate about a fixed horizontal axis perpendicular to its plane and passing through its

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centre. A string is wrapped over its rim and a block of mass m is attached to the free end of the string. The block is released from rest. If string does not slip on the rim then find the acceleration of the block. Neglect the mass of the string.

Solution :

Since string does not slip on the disc hence tangential acceleration of the point on the rim which is in contact with the string is equal to the acceleration of the block. Let angular acceleration of the disc about axis be , hence acceleration of the block a = R F.B.D. of the Block F.B.D. of the Disc. T a T

mg



mg - T = mR, as a =  R . . (1)

Eliminating T from (1) and (2), mg 2mg  I  2 m  MR ,   =  mR   R  as I =

Torque on the disc is = Tension + Mg I = TR as mg = 0 Where I = M.I. of the disc about the axis.  T = I/R . . . (2)

MR 2 2

Hence a =

3.7

Mg

2mg . 2m  M

Angular impulse

The angular impulse delivered by a torque in a given time interval is equal to the change in angular momentum. If angular momentum of a body is changed by a torque   during the period from a time t1 to a time t2, then  t2   J    dt where J = angular impulse. t1

  J 

t2



t1



 L2  dL dt  dL dt





L1



 J  L 2  L1  L

4.

Combined rotation and translation

In this type of motion, the axis of rotation is not stationary. If a body rotates about an axis with angular velocity  then the linear velocity of any particle located at a distance r from the axis of rotation, measured with respect to the axis of rotation    v   r

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 If the axis of rotation also moves with velocity v o with respect to a frame of reference, then the velocity of the particle relative to that frame will be     v    r  vo

4.1

Rolling Motion

 If velocity of the surface on which a body is rolling is v s (with respect to a frame) and R is the radius of the body, which is rotating with an angular velocity  around an axis, then the condition for pure rolling is,       R  vo  v s

 where v o is the velocity of the axis with respect to the frame.  If v s  0 then the condition for rolling is vo =  R L

Illustration 10:

A cylindrical drum, pushed along by a board rolls forward on the ground. There is no slipping at any contact. Find the distance moved by the man who is pushing the board, when axis of the cylinder covers a distance L.

Solution :

Let vo be the linear speed of the axis of the cylinder and  be its angular speed about the axis. As it does not slip on the ground hence 

vo . Where R is the radius of the cylinder. R

v = v o + R Speed of the topmost point is v = vo + R = 2vo Since time taken by the axis to move a distance L is equal to t = L/v o. In the same interval of time distance moved by the topmost point is v = v o  R L  2L S = 2vo  vo As there is no slipping between any point of contact hence distance moved by the man is 2L.

4.2

Energy of a rolling body

The translational K.E. of a rolling body 1  K.E. t  mv 2 2 and, the rotational K.E. of a rolling body 1  K.E.r  I 0  2 2



O

v

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Since we know that I 0  mk 2 and for rolling  



K.E.r 

 1 mk 2   2 

v r

v 2  1 k2  mv 2 2 r2 r 2 

...(b)

Therefore, the total K.E. of the rolling body = K.E. = K.E. t + K.E.r Putting v  r  for rolling, we obtain the total K.E. in terms of  as  1 k2  1  K.E.  m (r  ) 2  1 2  K.E.  (mr 2  mk 2 )  2 2 2 r   2 Since mk = M.I. of a body about its center of mass O  I O



mk 2 mr 2 I0 mr 2 IP where IP M.I.

of

the body about P. Therefore we conclude that, the combined effects of translation of center of mass of a body and its rotation about an axis passing through the c.m. are equivalent to its pure rotation about an axis passing through the point of contact P, of the rolling body. K. E. t % energy of translation =  100 K. E.

1 mv  2   12 mv  1 kr 2

2



2



r2 100  100 k 2 r 2

2



 

similarly, % energy of rotation 

k2  100 k2 r2

Illustration 11:

A thin hollow sphere of mass m = 2kg, radius = 1/2 m rolling on a horizontal surface with a constant frequency n = 60 rpm. Find the total mechanical energy of the hollow sphere.

Solution :

Translation K.E.  K.E. t 

1 mv 2 2 1 Rotational K.E.  K.Er  I2 2  Total energy 1 1 E  K.E.t  K.E.r  mv 2  I 2 2 2 2 v r , & I  mr 2 For rolling we 3 obtain 1 1 2  E m(r )2  mr 2  2   2 2 3  1 2 5  E  mr 2  2  1   mr 2  2   2 3 6

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Exercise 7 :

4.3



E  ( 5 / 6)mr 2 (2  n) 2 



E



E = 16.35 J.

10 m  2r 2n 2 3

10  1  ( 2 kg) (3.14) 2  m  2  3

2



60     60 sec 

2

Referring to the previous illustration, (a) what fraction of energy is translational? (b) what is the ratio of translation & rotational K.E.?

Role of friction in rolling motion

If a body rolls without slipping such that the velocity of its centre of mass does not change then no frictional force acts on the body. However, If a force acts on the body, the velocity of its centre of mass or its angular velocity about the centre changes and, there is a tendency of the body to slide at its point of contact. Thus, friction force acts on the body to oppose this tendency to slide.

4.4

Rolling on inclined plane

METHOD - I (Force- torque method) When a body rolls without slipping along an inclined plane, there are three forces acting on the body (1) mg - weight (2)

N - Normal reaction

(3)

f – static friction

Force Equation : Suppose that the body accelerates down the plane with an acceleration a.  The net force parallel to the plane

Fx  (mgSin   f )  ma

...(i)

Since the body does not move perpendicular to the plane, the net force acting on it along the perpendicular is zero. Fy  N  mg cos   0  ...(ii) Torque Equation:   Since mg & N pass through the centre O of the rolling body, torque about O due to these is zero. The torque due to friction, clockwise about O is   fr sin90

0

 fr

We know that  = Icm,  = Io



fr = Io

...(iii)

Constraint :

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Pure rolling implies that the point of contact P does not slide. It means that both velocity as well as acceleration of P relative to the plane must be zero, i.e. v P=0 and ap= 0.    We know that, a P  a O  a PO



aP = [ao  aPO] putting aP = 0



aP = 0, we obtain 0 = aO  aPO  aPO = aO r  a ...(iv)

a from (iv) and Io = mk2 in (iii) r a we obtain fr  mk 2 . r



Putting  =



mk 2 f 2 a r

...(v)

O

aPO = r P

a a0 = a

Acceleration : By putting the value of f from (v) in (i) we obtain,

mk 2 mg sin   2 a  ma r



a 

gsin 2



 1k   r 2  

Minimum friction required to sustain rolling : Putting the obtained value of a in (v) we get, mg sin  f  r2 1 k2 mgsin f   r2    1   k 2   where f = the minimum friction required to sustain rolling. The maximum value of static friction fmax   N. Putting N = mg cos, we obtain f max = mgcos Minimum coefficient of friction for sustained rolling

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Since f  fmax , putting the value of f & fmax mg sin    mg cos  2  we obtain  r  2  1  k 

tan 

 

r2   1 2 k  

tan 









 1 

r    k2  2

 min 

tan  r2 1 2 k

If    min , sliding occurs along with rotation. METHOD - 2 ( Energy method) We see that, the point of contact P does not slide in pure rolling. That means the point of application of frictional force f, does not move. Therefore, the work done by the friction is zero in pure rolling. That is why the total mechanical energy of the rolling body remains constant. The total energy of the rolling body between two positions 1 & 2, remains conserved. We obtain, ( PE) 1  2  ( KE) 1  2  0

 

 1 k2   mgh  mv 2  1 2   0 2 r  

v

1 h

2gh k2 1 2 r



h

2 

(  K.E) 1  2  K.E. 2  K.E.1



a v

 1 k mv 2  1 2 2 r 



2







 0

1 2  k2  mv  1  2  2 r   Using linear kinematics, We obtain v 22  v 12  2 a 

putting

v2  v 

Illustration 12 :

2gh  & h  sin  , we obtain a. k2 1 2 r A uniform disc of mass m and radius R rolls without slipping up a rough inclined plane at an angle of 300 with the horizontal. If the coefficient of static and kinetic friction are each equal to  and the only force acting on the disc are gravity



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and friction, then find the direction and magnitude of the friction force acting on it.

Solution :

Since disc does not slip friction is F.B.D. of the disc. N static and static friction can have any value between 0 and N.  Component of mg parallel to the plane is mgsin which is opposite to  the direction of motion of the centre mgsin of the disc, and hence speed of the mgcos centre of mass decreases. For pure  mg rolling the relation vc.m.= R must be obeyed. Therefore must decrease. Only friction can provide a torque about the centre.  Torque due to friction must be opposite to the  . Therefore frictional force will act up the plane. Now, for translational motion mg sin- f = mac.m. . . . .(i) For rotational motion fR = I, where I = M.I. of the disc about centre. a = I cm, as a = R R 

ac.m. =

fR 2 I

. . . (ii)

From (i) and (ii) we get, mg sin 

=

1

mR 2 I

Putting the value of and I we get  = mg/6 Illustration 13 :

A solid cylinder of mass m and radius r starts rolling down an inclined plane of inclination . Friction is sufficient to prevent slipping. Find the speed of its centre of mass when, its centre of mass has fallen a height h.

Solution :

Considering the two shown positions of the cylinder. As it does not slip hence total mechanical energy will be conserved. Energy at position 1 is E1 = mgh Energy at position 2 is E2 =

1 h 

2

1 1 mv c2.m.  Ic.m.2 2 2

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

2 v c.m.   , and Ic.m. = mr 2 r

3 mv c2.m. 4 From COE, E1 = E2 4  vc.m. = gh 3





E2 =

Note: In the previous example we used conservation principles, while in the one before we used Newton’s laws. Either one leads to the correct result, it is only a matter of convenience as to which method we choose. Conservation of angular momentum also helps in tackling problems concerning collisions of rolling bodies. Applying the principle of conservation of angular momentum about the point of collision helps to eliminate the external torques due to large impulsive forces.

Illustration 14 :

Solution :

A rigid rod of mass m & length  is pivoted at one of its ends. If it is released from its horizontal position, find the speed of the center of mass of the rod when it becomes vertical. 1 Suppose that in the vertical P

position the speed of the c.m. of the rod is v v  . OP    v 2 v ...(i) / 2  the total K.E. of the rod about the point P is given by, 1 K.E.  IP  2 ...(ii) 2



IO 

m 2 2  m  / 2 12



IO 

m 2 3



h = /2

v

O 2

where IP  I O  m(OP) 2

...(iii)

Using (b) & (c) K.E. 

1  m 2  2    2 3 

K.E. 

m 2  2 6

...(iv)

Using (a) & (b) K.E. 

m 2  2v    6  

2



2 mv 2 3

...(v)

Since the rod is released from rest, its K.E. at the position 1

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I  K. E.1  0 The change in K.E. = between the position 1 and 2 =  K.E. = K.E2 – K.E1 

K.E. =

2 2 mv 2  0  mv 2 3 3

. . . (vi)

Since the rod falls through a vertical distance =

 , its gravitational 2

potential energy decrease, change in P.E. is mg

 v

2  mv 2  mg 3 2 3g 4



 2



3 g 2

Illustration 15:

A plank of mass M rests on a smooth horizontal plane. A sphere of mass m is placed on the rough upper surface of the plank and the plank is suddenly given a velocity v in the direction of its length. Find the time after which the sphere begins pure rolling, if the coefficient of friction between the plank and the sphere is  and the plank is sufficiently long.

Solution :

Let t be the time after which slipping between the sphere and plank disappears. For the sphere, N = mg, N = mas  as = g 2  = I  mgr = mr2  5 5g 2r For the plank, mg N = Map  a p = M After time t, mg vp = v t velocity of plank, M velocity of sphere, vs = gt 5g t angular velocity of sphere,  = 2r For no slipping, the point of contact of sphere should have same speed as that



as mg

N

N N Sphere

vs

ap

N

Plank

vp

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of plank,  vs + r = vp Substituting and solving, mg  5g  gt +  t r = v t M  2r 



4.5

t=



2v

g 2 + 5 + 2 m



M



2v g[7  (2m / M)]

Instantaneous axis of rotation

The instantaneous axis of rotation is perpendicular to the plane of rotation and passes through the point which is momentarily at rest. About the instantaneous axis of rotation the combined motion can be treated as pure rotational motion. A perpendicular drawn to the axis of rotation from any point in the body is also perpendicular to the velocity of that point. Angular velocity of the body about instantaneous axis is same as that about its centre. Exercise 8:

A light rigid rod of length 1m is constrained to move in a vertical plane, so that its ends are along the x and y axes respectively. Find the instantaneous axis of rotation of the rod when it makes an angle  with horizontal.

Angular momentum of a body in combined translational and rotational motion Suppose that a body is rotating about an axis passing through its centre of mass with an angular velocity cm and translates with a linear velocity v. Then the angular momentum of the body measured about a point P outside the body in the lab frame is given by,     L P  L cm  r  p cm

where r is the position vector of the centre of mass with respect to point P. Hence,     L P  Icm   r  mv cm

Exercise 9:

A sphere rolls without slipping on a rough surface with centre of mass having constant speed v0. If mass of the sphere is m and its radius be R, then find the angular momentum of the sphere about the point of contact.

Exercise 10:

Shown in the figure is a rod which moves with v = 2m/sec & rotates with   2  rad / sec` . Find the instantaneous axis of rotation.

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5.

Solutions to the Exercises

Solution 1:

=

(i)

d  d =  dt dt 





t

d =

o



(ii)

0

 (4at

3

 3bt 2 ) dt

0

 = 0 + at

4

 bt 3

Further, =

d  d =  dt dt







t

t





d =  dt  (0 + at 4 - bt 3 ) dt

0

  = 0 t +

Solution 2 :



t

 dt =

0

5

o

4

at bt 5 4

A

Consider a thin elemental ring of thickness 'dr' at a radial distance 'r'. Mass of the disc = M M Mass per unit area of the disc =  R2

dr r

A

Area of the ring = 2r dr Mass of the ring =  (2r dr) Moment of the ratio of the ring = Mass (r)2

Solution 3:





  R

0

R

=



Moment of inertia of disc =

O

  2 dr  dr 2





 R4 2



MR 2 2



   r  F , where F = net force. 



In uniform circular motion r and F are collinear (but oppositely directed) 

  r F  0

   L  r p

 as linear momentum  p

is along the tangent,

P

hence   ˆ , r  p  rpn

where

n unit vector perpendicular

O

r

to the plane of the circle. 

Solution 4:

L  mvr  m r ,  Using the work energy theorem 2

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W= = Solution 5:

1 1  Fr  t (mr2)2 = mr 2  2 2  mr 2 

2

1 F2 t 2 = 0.56 N-m. 2 m

Since the gravitational force  GMm F aˆr r2  The torque of the force F about the sun = 

    r 



  

GMm r2

 aˆr  

 

1 mv2 2 1 2  mr 2  ( 2 ) KErotation =  2 3 

Solution 7:

  

  r F

GMm  ( r  aˆr )  0 r2

KEtranslation =

(v = r), KErotation =

1 mv2 3

5  1 1   mv 2  mv 2 6  2 3



KEtotal = 



KE trans 1/ 2 mv 2 3   2 KE total 5 5 / 6 mv KErotation 1/ 3 mv 2 2   . 2 KE total 5 5 / 6 mv

Solution 8:

Solution 9:

 If we drop perpendiculars upon v 1  and v 2 , they meet at 0. Therefore 0 is the instantaneous point of rotation and the axis passing through is perpendicular to the plane of rotation is known as axis of rotation. The coordinates of 0 is given as  cos  and  sin  respectively. 

O v1 

O

v2

x



      Icm  R  m v cm

Since LP  L cm  r  pcm

Since sphere is in pure rolling motion hence  = v0/R  Lp = Solution 10:

y

v0 P

2 v 7 MR2 0  Mv 0R  Mv 0R 5 R 5

Let P be a point on the instantaneous axis of rotation.

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Therefore v P  0 .   | v PO  v O |  0 

 v PO  v O 0



v PO  v O where v O  v & v PO  (OP) 



( OP ) v



OP  v /  

2 1  m. 2.  



O P

r

v = 2 m/s v

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