# Career Point Sample_1.pdf

September 4, 2017 | Author: Shalini Kureel | Category: Acceleration, Force, Length, Geometric Measurement, Nature

#### Description

CIRCULAR MOTION

IIT-JEE Syllabus

1. Kinematics in two dimension 2. Circular motion

Total No. of questions in Circular motion are: Solved examples…....…………………………..…15 Exercise # 1 …….……………………………….…20 Exercise # 2 …….……………………………….…29 Exercise # 3 …….……………………………….…31 Exercise # 4 ……………………………………..…07 Exercise # 5 ……………………………………..…12 Total No. of questions………………..114

*** Students are advised to solve the questions of exercises in the same sequence or as directed by the faculty members. CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

5

Index : Preparing your own list of Important/Difficult Questions Instruction to fill (A) Write down the Question Number you are unable to solve in column A below, by Pen. (B) After discussing the Questions written in column A with faculties, strike off them in the manner so that you can see at the time of Revision also, to solve these questions again. (C) Write down the Question Number you feel are important or good in the column B.

EXERCISE NO.

COLUMN :A

COLUMN :B

Questions i am unable to solve in first attempt

Good/Important questions

1

2

3

4

5

Advantages 1. It is advised to the students that they should prepare a question bank for the revision as it is very difficult to solve all the questions at the time of revision. CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

6

2. Using above index you can prepare and maintain the questions for your revision.

KEY CONCEPT (ii) Instantaneous Angular velocity :

1. Rotational kinematics

 = lim

1.1 Angular Displacement

t 0

Introduction : Angle substended by the position vector of a particle moving along any arbitrary path w.r.t. some fixed point is called angular displacement.

2. Relation between linear velocity and angular velocity

Q

v  r

 S

O r

d ds = linear velocity and  = dt dt    In vector form, v    r

v=

P

Particle moving in circular path

angle 

Note : (i)

 d = t dt

 v

360º  radian = 180º 2

 O 

If a body makes n revolutions, its angular displacement  = 2n radians

 r

  Outward normal to plane of paper. O  (direction of  )

1.2 Angular Velocity It is defined as the rate of change of angular displacement of a body or particle moving in circular path.

(ii) When a particle moves along a curved path, its linear velocity at a point is along the tangent drawn at that point

Its direction is same as that of angular displacement

(iii) When a particle moves along curved path, its velocity has two components. One along the radius, which increases or decreases the radius and another one perpendicular to the radius, which makes the particle to revolve about the point of observation.

i.e. perpendicular to plane of rotation Note : If the particle is revolving in the clockwise direction then the direction of angular velocity is perpendicular to the plane downwards. Whereas in case of anticlockwise direction the direction will be upwards. (i) Average Angular Velocity : av 

T otalangular displacement T otaltime taken

3. Angular acceleration (i) The rate of change of angular velocity is defined as angular acceleration.     d   lim  t 0 t dt

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

7

(ii) Its direction is that of change in angular velocity

1   0 t   t 2 2

4. Equations related to motion of a particle with constant angular acceleration

2 = 02 + 2 

  0   t

5. Equation of linear motion and rotational motion S.N.

Linear Motion

Rotational Motion

(i)

With constant velocity

a = 0, s = ut

 = 0 ,  = t

(ii)

With constant acceleration

(i) Average velocity

(i) Average angular velocity

vav =

vu 2

av =

(ii) Average acceleration aav =

vu t vu t 2

(iii)  = av. t =

(v) s = ut +

1 2 at 2

(v)  = 0t +

(vi) s = ut –

1 2 at 2

(vi)  = 0t –

(vii) v2 = u2 + 2as (viii) Sn = u +

1 (2n – 1)a 2

displacement in nth sec. With variable acceleration

2  1 t

(i) v =

1  2 t 2

(iv)  = 0 + t

(iv) v = u + at

(iii)

(ii)Average angular acceleration av =

(iii) s = vav t =

1  2 2

ds dt

(ii)  ds =  vdt dv dv (iii) a = =v dt ds

(iv)  dv =  a dt; if a = f(t)

1 2 t 2 1 2 t 2

(vi) 2 = 02 + 2 1 (2n–1) 2

(viii) n = 0 +

Angular displacement in nth sec (i)  = d/dt (ii)  d =  dt (iii)  =

d d = dt d

(iv)  d =   dt ; if  = f(t) (v)  d =  d; if  = f()

(v)  vdv =  a ds; if a = f(s)

6. Acceleration in circular motion

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

8

y

eˆ t

eˆ r

P  O

x

eˆ r : unit vector along the outward radius eˆ t : unit vector along the tangent in the direction of increasing .

(i) The velocity of the particle changes while moving on the circular path, this change in velocity is brought by a force known as centripetal force and the acceleration so produced in the body is known as centripetal acceleration. (ii) The direction of centripetal force or acceleration is always towards the centre of circular path. (iii) Expression for centripetal acceleration:

eˆ r = ˆi (cos)  ˆj(sin ) and

 ac =

v2 = r2 r

eˆ t =  ˆi (sin )  ˆj(cos)

 v

where ˆi and ˆj are the unit vectors along x and y axes respectively.

O

velocity of particle    dr , where r = r(ˆi cos  ˆj sin ) v = dt r is the radius of circle

P

 ac

 ac

  v  v

 ac

  

(iv) The direction of a c would be the same as 

that of  v (change in velocity vector) (v) Expression for Centripetal force : If v = velocity of particle , r = radius of curvature of path Then necessary centripetal force Fc = mass × acceleration

It is defined as the rate of change of speed.   d| v | | at | = dt

Fc = m

v2 r

Its direction is along the tangent to the path. 

 v

   at =  r

 FC

  d where  = dt

7. Centripetal acceleration and centripetal force

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

  represents  direction of 

This is the magnitude of centripetal acceleration of particle. It is a vector quantity. In vector form

acceleration of particle :   dv a = dt   dv  a = (2 r )eˆ r    eˆ t  dt     a = ac  at  where a c (centripetal acceleration) = 2r( eˆ r )  dv and (eˆ t ) a t (tangential acceleration) = dt  Tangential Acceleration (a t )

 r

 v

Note : 

Centripetal force is not a real force. It is only the requirement for circular motion.

CIRCULAR MOTION

9

 v2

It is not a new kind of force. Any of the forces found in nature such as gravitational force, electrostatic force friction force, tension in string, reaction force etc may act as centripetal force.

 v1

 v3

8. Centrifugal force 

It is a sufficient pseudo force, only if we are anlaysing the particles at rest in a uniformly rotating frame.

(ii) As | v | is constt. so tangential acceleration at = 0

  ac

m

O

at=0 (iii) Tangential force Ft = 0

r In the given figure, the block of mass 'm' is at rest with respect to the rotating platform (as observed by the observer O on the rotating platform). N 2

mr fs

 Fc

Ft=0 (iv) Total acceleration a = a c2  a t 2 = ac= (towards the centre)

mg

centrifugal force = m r2 centrifugal force acts (or is assumed to act) because we describe the particle from a rotating frame which is non-inertial and still use Newton's laws.

(v) In uniform circular motion Ft =0, as at = 0, so work done will be zero by tangential force. But in non-uniform circular motion Ft  0, thus there will be a work done by tangential force in this case. Rate of work done by net force in nonuniform circular motion = rate of work done by tangential force

9. Type of circular motion 9.1 Uniform circular motion 9.2 Non Uniform Circular Motion :

 P=

9.1 Uniform Circular Motion : If m = mass of body , r = radius of circular orbit, v = magnitude of velocity ac = centripetal acceleration, at = tangential acceleration In uniform circular motion : 

(i) | v1 | = | v 2 | = | v 3 | = constant

v2 r

  dW = Ft . v dt

Note:  Because Fc is always perpendicular to velocity or displacement, hence the work done by this force will always be zero.  There is an important difference between the projectile motion and circular motion. In projectile motion, both the magnitude and the direction of acceleration (g) remain constant, while in circular motion the magnitude remains constant but the direction continuously changes.

i.e. speed is constant CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

10

9.1.1 Motion In Horizontal Circle : Conical pendulum  F  Fc 

S  T T cos T sin O OP=r



(vi) Net acceleration, P

tan  =

v4 r 2g 2

at F = t ac Fc

....(i)

v2 and tan = rg

 a    a c  ac  at

....(ii)

9.2.1 Motion in Vertical Circle : Motion of a body suspended by string

9.2 Non-uniform Circular Motion : (i) In non-uniform circular motion :

When the body rises from the bottom to the height h, a part of its kinetic energy converts into potential energy

| v |  constant   constant i.e. speed  constant

(a) Velocity at a point P :

i.e. angular velocity  constant

dv (ii) Tangential acceleration : at = , dt

where v =

Fnet m

The angle made by 'a' with ac,

and T cos = mg

From these equation T = mg 1 

a c2  a 2t =

a=

mg

mv2 r

 T sin=

 Fc  Ft 

ds and s = arc length dt

Total mechanical energy remains conserved at point A and point P.



0+

v=

(iii) Tangential force : Ft = mat mv 2 (iv) Centripetal force : Fc = = m2r r

1 1 mu2 = mgh + mv2 2 2

u 2  2g(1  cos) ;

u 2  2gh =

as h =  –  cos  [Where  is length of the string]

(v) Net force on the particle : 

F  Fc  Ft 

F=

Fc2  Ft 2

F tan = t Fc

O

v  

h

P u

A

(b) Tension at a point P : At point P required centripetal force mv2 =  CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

11

3g

The critical velocity at C = v

O h

Also TA = 6mg, TB = 0 , TC = 3mg

 T P  mgcos

(iii) Condition of leaving the circular path 2g < u <

mg

A mgsin

In this case particle will not follow circular motion. Tension in string becomes zero somewhere between points C & B whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory

Net force towards the centre = T – mg cos This net force provides required centripetal force. 

mv 2 

 T – mg cos =

5g

B v

2

T = m [g cos + T=

v ] 

 D

m 2 [u – gl (2–3cos)] 

...(1)

(c) Tangential force for the motion

A

Ft = mgsin

(iv)Condition of oscillation :

This force retards the motion.

0

C



(a) If u =

5g

In this case tension in the string will not be zero at any of the point, which implies that the particle will continue the circular motion. (ii) Condition of looping the loop : u =

5g

(b) If u <

2g

The velocity of particle will become zero between A and C but tension will not be zero and the particle will oscillate about the point A.

Critical Velocity : The minimum velocity at which the circular motion is possible

v

2g

In this case both velocity and tension in the string becomes zero at point C and D and the particle will oscillate along semicircular path.

In this case the tension at the top most point (B) will be zero, which implies that the particle will just complete the circular motion.

B

2g 

10. Dynamics of circular motion in horizontal plane  Maximum speed of vehicle for safe turning on rough horizontal circular turn

C

vmax =

A 5g

µs Rg

where

The critical velocity at A =

5g

The critical velocity at B =

g

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

R  radius of curvature of circular turn µs  Coefficient of static friction

CIRCULAR MOTION

12

N O

R

fs

mg

fs  µsmg fs =

mv 2  R

Static friction provides the required centripetal force.

N O

tan  = v=

Nsin 

R 

Ncos



mg



mg

v2 Rg

Rg tan

v  safe speed on banked road.

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

13

SOLVED EXAMPLES Ex.1

Sol.

The magnitude of the linear acceleration of the particle moving in a circle of radius of 10cm with uniform speed completing the circle in 4s, will be The distance covered in completing the circle is 2r = 2 × 10cm The linear speed is 2 10 2r v= = = 5 cm/s 4 t The linear acceleration is, (5) 2 v2 = = 2.52 cm/s2 10 r This acceleration is directed towards the centre of the circle

Substituting the given values, we have a = 4 × (3. 14)2 × (5.3 × 10–11) (6.6 × 1015)2 = 9.1 × 1022 m/s2 towards the nucleus. The centripetal force is FC = ma = (9.1 × 10–31) (9.1 × 1022) Ex.4

a=

Ex.2

Sol.

Sol.

The length of second's hand in a watch is 1cm. The change in velocity of its tip in 15 seconds is Velocity = =

ac =

Circumfere nce Time of revolution

2  1  2r = = cm/s 60 30 60

Change in velocity v = =

         30   30   30

2

2 cm/s

An electron is moving in a circular orbit of radius 5.3 × 10–11 metre around the atomic nucleus at a rate of 6.6 × 1015 revolutions per second. The acceleration of the electron and centripetal force acting on it. will be - (The mass of the electron is 9.1 × 10–31kg.) Let the radius of the orbit be r and the number of revolutions per second be n. Then the velocity of electron is given by v = 2nr,   acceleration a =

Write an expression for the position vector r for a particle describing uniform circular motion, using rectangular coordinates and the unit vectors i and j. The vector expressions for the velocity v and acceleration a will be 

Sol.

Sol.

250 250 v2 = = 62.5 m/s2 1000 r a Centripetal accelerati on 62.5 = c = gravitational accelerati on 9 .8 g

= 6.38 : 1 Ex.5

2

Ex.3

= 8.3 × 10–8 N towards the nucleus. An air craft executes a horizontal loop of radius 1km with a steady speed of 900km/h. The ratio of centripetal acceleration to that gravitational acceleration will beGiven that radius of horizontal loop r = 1 km = 1000 m 900 5 Speed v = 900 km/h = = 250 m/s 18 Centripetal acceleration

r = ˆi x + ˆj y, x = r cos, y = r sin 

where  = t 

r = ˆi ( r cos t) + ˆj (r sin t) 

v = d r /dt = – ˆi (r sin t) + ˆj (r cos t) 

a = d2 r /dt2 = –2 r Ex.6

4 2 r 2 n 2 v2 = r r = 4 2 r n2

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

The vertical section of a road over a canal bridge in the direction of its length is in the form of circle of radius 8.9 metre. Find the greatest speed at which the car can cross this bridge without losing contact with the road at its highest point, the center of gravity of the CIRCULAR MOTION

14

Sol.

O

car being at a height h = 1.1 metre from the ground. (Take g = 10m/sec2) Let R be the normal reaction exerted by the road on the car. At the highest point, we have



Ex.7

Sol.

=

Ex.8

Sol.

mv r

2gr or v2 = 5

v2 =

cos  = (4/8) = 

mv  sin

...... (B)

1 ,  = 60º 2

The angle with horizontal = 90º – 60º = 30º

 2g    = 7.8 × 104 radian/sec.  5r 

A man whirls a stone round his head on the end of a string 4.0metre long. Can the string be in a horizontal, plane? If the stone has a mass of 0.4kg and the string will break, if the tension in it exceeds 8N. The smallest angle the string can make with the horizontal and the speed of the stone will respectively be (Take g = 10m/sec2.)

=

mg cos When the string is horizontal,  must be 90º i.e.,cos 90º = 0 mg  T= = 0 Thus the tension must be infinite which is impossible, so the string can not be in horizontal plane. The maximum angle  is given by the breaking tension of the string in the equation T cos  = m.g. Here T (Maximum) = 8N and m = 0.4 Kg.  8 cos  = 0.4 × g = 0.4 × 10 = 4

mv 2 3 mg=mg– 5 r

mv 2 2 mg= 5 r

...... (A) 2

2

Form eq. (A) T =

mv 2 3 W=mg– 5 r

2  9.8  (6400103 ) 5 Solving, we get v = 5 × 109 m/sec,

Now

T sin  =

much as present will be (Take the equatorial radius as 6400km) Let v be the speed of earth's rotation. We know that W = mg

or

Tsin mg mg

T cos = mg

The angular speed with which the earth would have to rotate on it axis so that a person on the equator would weigh (3/5)th as

Hence

r A

Therefore v2  (r + h)g = (8.9 + 1.1) × 10 or v2  10 × 10  v  10  vmax = 10 m/sec

Tcos 

T

T

mv 2 = mg – R, R should not be negative. (r  h )



From equation (B), 8 sin 60º =

0.4  v 2 4 sin 60o

32sin 2 60º v2 = = 80 sin2 60º  0.4

 v=

Ex.9

A smooth table is placed horizontally and a

Form figure Sol.

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

80 sin 60º = 7.7 m/sec.

spring of unstreched length l0 and force constant k has one end fixed to its centre. To the other end of the spring is attached a mass m which is making n revolutions per second around the centre. Tension in the spring will be. Let T be the tension produced in the stretched string. The centripetal force required for the mass m to move in a circle is provided by the CIRCULAR MOTION

15

(a) From figure h = l (cos – cos0)

tension T. The stretched length of the spring is r (radius of the circle). Now,

and v2 = 2 g h = 2 g l (cos – cos 0) .... (A)

Elongation produced in the spring = (r – l0)

Again T – mg cos  = mv2/l

Tension produced in the spring, T = k (r – l0)

Substituting the value of v2 from eq. (A) in eq. (B) we get

........(A)

Where k is the force constant Linear velocity of the motion v = 2r n. m(2rn) mv = r r 2

 Centripetal force =

T – m g cos  = m {2 g l (cos – cos 0)/l}

2

or T = m g cos  + 2 mg (cos – cos 0) or T = 40 g (3cos – 2 cos 0) newton

= 42 r n2 m ........(B) Equating equation. (A) and (B), we get

or T = 40 (3cos – 2 cos 0) kg f.

k (r – l0) = 42 r n2 m ( T = mv2/r) 

(b) Let 0 be the maximum amplitude. The

 kr – k l0 = 4 2 r n2 m

maximum tension T will be at mean position where  = 0  Tmax = 40 (3 – 2 cos 0),

r (k – 42 n2 m) = k l0 k 0

r=

( k  4 n m)

  k 0 T=k   0  2 2  ( k  4 n m ) 

or T =

But Tmax = 80 kgf

........(C)

2 2

Substituting the value of r in eqn. (A) we have

4 2 n 2 m 0 k

........ (D)

( k  4 2 n 2 m)

Solving we get  = 60º Ex.11 An aircraft loops the loop of radius R = 500 m with a constant velocity v = 360 km/hour. The weight of the flyer of mass m = 70 kg in the lower, upper and middle points of the loop will respectively beSol. See fig, Here v = 360 km/hr = 100 m/sec.

N

Ex.10 A 40 kg mass, hanging at the end of a rope of

mg

length l, oscillates in a vertical plane with an

N

angular amplitude of 0. What is the tension

Sol.

.... (B)

in the rope, when it makes an angle  with the vertical? If the breaking strength of the rope is 80 kgf, what is the maximum angular amplitude with which the mass can oscillate without the rope breaking? The situation is shown in fig  0



N mg At lower point, N – mg =

N = weight of the flyer = mg + N = 70 × 10 +

T

h

mv2 , R

70 (10000) = 2100N 500

At upper point, N + mg =

v

2  mv  

mg

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

N=

mv2 . R

mv2 , R

mv2 – mg = 1400 – 700 = 700N R CIRCULAR MOTION

16

At middle point, N =

Ex.12 A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along, the wall at a distance 3 m. What is the velocity of the spot Sol.

P when  = 45º If x is the distance of point P from O then from fig. S

 3m P



O

wall

////////////////////////////////////////////////////////

or or

(E –p) T = 2

[as  = t]

or

 2 2  2    = T  TE TP 

2   as T     

or

1 1 1 = – , T TE TP

i.e., T =

TP TE TP  TE

So

T=

mv2 = 1400N R

x tan  = (x/h) x = h tan  dx d = h (sec2 ) dt dt

i.e. v = h sec2    [as (dx/dt) = v and (d/dt) = ] Here h = 3 m,  = 180 – (45 + 90) = 45º and = 0.1 rad/s. So v = 3 × (2)2 × 0.1 = 0.6 m/s.

3 1 = 1.5 year 3 1

Ex.14 A small object loops a vertical loop from which a symmetrical section of angle 2 has been removed as shown in fig. Find the maximum and the minimum heights from which the object, after losing contact with the loop at point A and flying through air, will reach point B. Find the corresponding angles of the section removed for which this is possible– Sol. In order that the particle may fly off from A and land at B, the range of the particle must be –

V0 B

A 2 O

Ex.13 A planet P revolves around the sun in a circular orbit, with the sun at the centre, which is coplanar and concentric to the circular orbit of earth E around the sun. P and E revolve in the same direction. The time required for the revolution of P and E around the sun are 3 years and 1 year respectively. What is the time required for P to make one revolution around the sun relative to E Sol. As TP > TE and T = 2/ so p < E and hence with respect to sun the difference in their angular displacement per unit time will be (E –p). So they will be at same position with respect to the sun again for the first time when their relative angular displacement becomes 2. So if T is the required time CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

v 02 gR sin 2  v 02 = g cos Applying the law of conservation of energy 1 mgH = mg R (1 + cos ) + m v 02 2 1 gR  mgH = mgR (1 + cos ) + m 2 cos   2 cos2  – 2 (k – 1) cos  + 1 = 0 H where k = R

AB = 2R sin  =

2(k  1)  4(k  1) 2  8 2 2 1 1 (k  1) 2  2 = (k–1) ± 2 2 Since cos  is real (k – 1)2 2  k 1 + 2

 cos  =

CIRCULAR MOTION

17

0 < cos   1, 1 1 (k – 1) ± (k  1) 2  2 1 2 2  (k – 1)2 – 2 (3 – k) 2  k  2.5 Since

Thus 1 + 

2 k  2.5 

 (1 + 2 ) R  H 2.5 R  Hmax = 2.5 R and Hmin = (1 + When

2 ) R = 2.4 R

H = Hmin

and cos  =

1

k=1+ 2

  = 45º

ay =

For

y = ymin,

2

When

H = Hmax

and cos  = 1,

k=

5 2

1   = 0º,60º 2

The solution  = 0 is not acceptable as that will mean no cut,   = 60º for Hmax. Ex.15 A particle moves in a plane according to X = R sin t + R and y = R cos t + R Where and R are constant. This curve, called a cycloid, is the path traced out by a point on the rim of a wheel which rolls without slipping along the x-axis. Find the instantaneous velocity and acceleration when the particle is at its maximum & minimum value of y. dx dy Sol. = R cost and = – R sint dt dt ymin = 0, when t = 

when t = .

then

d2x =0 dt 2

and

d2y = R2 .  dt 2



then 

d2y = – R 2 cos t dt 2

and

a = R2 y = ymax' when t = 2, d2x d2y = 0 and = – R2 . dt 2 dt 2

a = – R2

dx dy = – R and = 0, dt dt v = – R ymax = 2R, when t = 2

Now

dx dy = R and =0 dt dt

Now v =

 dx  2  dy  2        = R  dt   dt  

ax =

d2y = – R2 sin t dt 2

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

CIRCULAR MOTION

18