Solutions to Concepts of Physics by HC Verma Chapter 10

10 1.

Rotational Mechanics A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds. With usual notation we have ⇒ ( ⇒

2.

3.

(

)

)

∫

A wheel rotating with a uniform angular acceleration covers 50 revolutions in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds. Here

,

⇒

(

,

, We also know that α is constant.

)

A wheel starting from rest is uniformly accelerated at for 10 seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in the next 10 seconds. Find the total angle rotated by the wheel. Here

,

,

,

,

,

,

and and ⇒ ⇒

4.

A body rotates about a fixed axis with an angular acceleration of one radian/second/second. Through what angle does it rotate during the time in which its angular velocity increases from 5 rad/s to 15 rad/s? Here

,

,

⇒ ⇒

5.

Find the angular velocity of a body rotating with an angular acceleration of comletes the 5th revolution after the start. Here

,

,

10.1

as it

10

Rotational Mechanics

⇒

√

√

∴

6.

7.

√

A disc of radius 10 cm is rotating about its axis at an angular speed of linear speed of (a) a point on the rim (b) the middle point of the radius. (a)

Here ⇒

.

,

(b)

Here ⇒

.

,

. Find the

A disc rotates about its axis with a constant angular acceleration of . Find the radial and tangential acceleration of a particle at a distance 1 cm from the axis at the end of the first second after the disc starts rotating. Here

.

,

,

⇒

. .

8.

A block hangs from a string wrapped on a disc of radius 20 cm free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is at some instant, with what speed is the block going down at that instant? The block will be going down at the translational speed of a point on the rim of the disc. ⇒ . ( )

9.

Three particles, each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertia of the system about an axis (a) joining two of the particles (b) passing through one of the particles and perpendicular to the plane of the particles. (a)

The two points on the axis, each having zero distance from the axis, contribute zero to the moment of inertia. The distance of the third point from the axis is . sin √ ∴ . ( ) . x

(b)

Here distance of the two points from the axis is 10 cm each. The third point makes zero contribution to moment of inertia because its distance from the axis is zero. ( . ) ⇒ . x

10. Particles of masses g, g, g, ……,

g are kept at the marks cm, cm, cm,……., 100 cm respectively on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. The distances of the g, g, …. g particles from the center and therefore the axis are 9 cm, cm, ……, cm. Similarly the distances of the g, g, …….., g particles are cm, cm, ……., cm. Thus

10.2

10

Rotational Mechanics ∑ ∑ ∑

(

)

∑

(

( .

) (

) 9

)(

)(

.

(

)

) (

(

)

)

.

11. Find the moment of inertia of a pair of spheres, each having a mass in contact about the tangent passing through the point of contact.

and radius , kept

Let and be the moment of inertia of the first and the second sphere about the given commom tangent. Then by applying the parallel axis theorem and noting that distance between the parallel axes is , we have ⇒

12. The moment of inertia of a uniform rod of mass 0.50 kg and length 1 m is 0.10 about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod. .

With usual notation we have ⇒

)

√(

√

.

. .

.

13. Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles.

Let be the mass of the circular ring and I the moment of inertia about a line perpendicular to the plane of the ring and passing through one of the particles of the ring. Then ⇒

√

√

√

14. The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the center. Here we have

⇒ ∴ ⇒

√

√

⇒ √

15. Find the moment of inertia of a uniform square plate of mass its diagonals.

and edge about one of

Let the moment of inertia about the axis through the center of the square plate perpendicular to the plane, taken as z-axis, be I. Let the two lines through the center of

10.3

10

Rotational Mechanics the square plate and parallel to the edges be the x-axis and the y-axis. Then by symmetry the moment of inertia about x-axis and y-axis is same. By perpendicular axis theorem, the moment of inertia about z-axis is the sum of moments of inertias about x-axis and y-axis. ⇒ Since the two diagonals of the square plate are perpendicular to each other, next let them be the x-axis and y-axis respectively. By perpendicular axis theorem, the moment of inertia about z-axis is the sum of moments of inertias about x-axis and y-axis. Thus ⇒

16. The surface density (mass/area) of a circular disc of radius depends on the distance from the cemter as ( ) . Find its moment of inertia about the line perpendicular to the plane of the disc through its center.

Take a circular strip of radial difference with center at the center of the disc. The area ) of the strip is ( . If is the mass of the strip then ( ) ( ) . The distance of this strip from the axis being the moment of inertia of this strip is ( ) Integrating over 0 to ∫

(

)

*

+

(

)

17. A particle of mass

is projected with a speed at an angle θ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point. At the highest point of the motion of the particle, the force parallel to x-axis is zero and the force parallel to y-axis is mg. The perpendicular moment arm is half the range of motion. Thus sin cos

18. A simple pendulum of length is pulled aside to make an angle θ with the verticle. Find the magnitude of the torque of the weight When is the torque zero?

of the bob about the point of suspension.

In this case the force is and is directed downward. As the initial position of the pendulum string makes an angle θ with the verticle and the length of the string is , the perpendicular moment arm is sin . Thus The above result shows that the torque is zero when θ is zero, that is when the bob is at the lowest point.

19. When a force of 6.0 N is exerted at 30° to a wrench at a distance 8 cm from the nut, it is

just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicular to the wrench at 16 cm from the nut?

10.4

10

Rotational Mechanics When 6.0 N force acts on the wrench at 30° at a distance 8 cm from the nut, the torque acting the nut is . ( . ) sin . . Thus the force required to loosen the nut when it is applied perpendicular to the wrench 16 cm from the nut will be given by ( . ) sin . ⇒

.

.

.

20. Calculate the total torque acting on the body shown in the figure about the point O. . . .

( ) ( ) sin .

(

. .

.

) sin ( ) sin . .

15 N 10 N 37° 6 cm 90°

4 cm

O 4 cm

3 cm 30° 5N

150° 20 N

21. A cubical block of mass

and edge slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center. The block moves down the incline with uniform speed and therefore the net force along the incline is zero. Thus sin ⇒ sin The gravitational force due to symmetry about the center θ produces no torque. Since the only forces acting on the block are the gravitational force , normal force N and the frictional force , the net torque is provided only by the later two forces. ∴ But the block does not roll and therefore the net torque is zero. ⇒ ( ) sin Thus the torque due to normal force is in the clock-wise direction.

22. A rod of mass m and length L, lying horizontally, is free to rotate about a verticle axis

through its enter. A horizontal force of constant magnitude F acts on the rod at a distance L/4 from the center. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after it starts. With usual notation we have sin 9 Noting that the moment of inertia of the rod through an axis parallel to it is zero, the perpendicular axis theorem gives us

10.5

10

Rotational Mechanics ( )x( ) If θ is the angle rotated in time , and initial angular velocity

being zero, we have

23. A square plate of mass 120 g and edge 5.0 cm rotates about one of the edges. If it has a uniform angular acceleration of .

, what torque acts on the plate?

By parallel axis theorem (

( ) ⇒

. x

)

( .

)(

)

.

.

24. Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with same angular acceleration.

Let be the moment of inertia of the square plate through it diagonal. Then is also the moment of inertia through the other diagonal. From the perpendicular axis theorem the moment of inertia through the axis passing through the center of the square plate and perpendicular to it is . By symetry of the square plate about the lines through its center and parallel to its edges is also the sum of moment of inertias through these two perpendicular lines. Thus (

)

⇒

( )(

⇒

x

)(

)

. x

.

.

25. A flywheel of moment of inertia .

is rotated at a speed 60 rad/s. Because of the friction of the axle, it come to rest in 5 minutes. Find (a) the average torque of friction, (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating. (a)

.

Here .

.

Since the fly wheel comes to rest in 5 minute we have ( ) ⇒ ⇒ . (b)

The work done by the frictional force is the change in energy of the fly wheel. The initial and final translational velocity of the fly wheel is zero. There is also no change in the potential energy. Therefore ( )( ) 9 ⇒

(c)

9. k

The angular momentum of the wheel is

10.6

10

Rotational Mechanics (

)

(

)

.

26. Because of the friction between the water in the oceans with the earth's surface, the

rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by 0.0016 rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is . kg. The average torque of friction is given by ( . x )( .

⇒

.

. x

. x

where )

9. x (

)

.

x

.

27. A wheel rotating at a speed of 600 rpm (revolutions per minute) about its axis is brought

to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque. Given that ⇒

. Let be the constant torque applied to stop the wheel. Then (

)

⇒

(

)

28. A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed 100 rev/min

when the motor is turned off. Neglecting the friction at the axle, find the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions. With usual notation .

.

Let α be the deceleration caused by the applied tangential force F. Then x( ⇒

. ( .

)

) .

29. A cylinder rotating at a speed of 50 rev/s is brought in contact with an identical

stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ? Let be the initial angular velocity of the first cylinder and be the initial angular velocity of the second one. Let the common angular acceleration of the two cylinders be say . . . Let be the common angular velocity after time . Then

10.7

10

Rotational Mechanics and ⇒ or

30. A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration

of . At what time will the body have kinetic energy same as the initial value if the torque continues to act? Let the constant torque be . Then ( (

)

(

)

(

)

) (

⇒

)

(

)

( (

(

)

)

)

31. A light rod of length 1 m is pivoted at the center and two masses of 5 kg and 2 kg are

hung from the ends as shown in the figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning. With usual notation ( ) . .

( )

2 kg

5 kg

. ⇒

.

( . )

( )

.

.

32. Suppose the rod of the previous problem has a mass of 1 kg distributed uniformly over its length. (a) Find the initial angular acceleration of the rod. (b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg. (a)

Let be the moment of inertia of the rod through the center without blocks. Then the moment of inertia of the rod/blocks system is ( )

( ) ( . )

x( ) (b)

Let

and

⇒ Similarly ⇒

. .

.

be the tensions in the 2 kg and 5 kg block supports. Then . 9. . . 9.

.

9

. ( . )

33. Figure shows two blocks of masses

.

and M connected by a string passing over a pulley. The horizontal table over which the mass slides is smooth. The pulley has a radius and moment of inertia I about its axis and it can freely rotate about this axis. Find the

10.8

10

Rotational Mechanics acceleration of the mass M assuming that the string does not slip on the pulley. Let

be the acceleration of the mass M. Then

Tm

m

From the motion of the pulley we have ( ) But . Putting values of , and we have ( )

TM

and solving

M

for

⇒

34. A string is wrapped on a wheel of moment of inertia .

. and radius 10 cm and goes through a light pulley to support a block of mass 2.0 kg as shown in the figure. Find the acceleration of the block. Let be the acceleration of the 2 kg block and the tension in the string on either side of the light pulley. From the motion of the block we have From the motion of the pulley of moment of inertia . have

.

we 2 kg

⇒

(

) . 9

.

.

.

35

Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia . . . Find the tension in the part of the string joining the pulleys. Let be the tension in the string joining the two pulleys and the tension in the string between the block and the pulley. Let and be the angular accelerations of the bigger and the smaller pulleys respectively. Then from the motion of the block ( ) Let and be the radii of the bigger and the smaller pulleys and I and I' be the moment of inertia of the bigger and the smaller pulleys respectively. From the motion of the bigger and smaller pulleys respectively ⇒ . ( ) and Also and equation we have * (

. Putting these values and those of T and T' in the above

)

+

⇒ ⇒

.

.

.

(

)

. .

.

10.9

10

Rotational Mechanics

36. The pulleys in the figure are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

Let the acceleration of the block M be . Let , and be the tensions as shown in the figure. As the two pulleys have same radii, they will have same angular acceleration say α. From the motions of the blocks M and m we have

T2 T1 M

T3 m

From the rotation of the two pulleys we get ( ) ( ) ⇒ ( )( ) (

⇒

)

37. The descending pulley shown in the figure has a radius 20 cm and moment of inertia . The fixed pulley is light and horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg. .

.

Let be the acceleration of the block. Let the tensions in the three T1 parts of the string be as shown in the figure. From the motion of 1 kg the block we get T2 T1 …I The descending pulley has two motions viz., the rotation and the translation. Let the angular acceleration of the pulley be α. Then from the rotation of the descending pulley, we have (

)

.

(

)

.

Clearly the downward acceleration of the descending pulley is ⇒

. Therefore

⇒ … II Let M be the mass of the descending pulley, then ( . ) ( . )

⇒

… III

From the downward motion of the descending pulley we have ( ) … IV Substituting values of , and in equation IV we get ( ) ⇒

.

.

.

38. The pulley shown in the figure has a radius 10 cm and moment of inertia .

. about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.

10.10

10

Rotational Mechanics Let the acceleration of the 4 kg block be Then motion of 4 kg block gives us sin …I The motion of 2 kg block gives sin … II Rotation of the pulley gives . ( )( . ) ⇒ Substituting values of , √ √ ⇒ . √

… III in III above

down the incline. T2

T1 4 kg

2 kg

45°

45°

.

39. Solve the previous problem if the friction coefficient between the 2.0 kg block and the plane below it is 0.5 and the plane below the 4.0 kg block is frictionless.

Let the acceleration of the 4 kg block be down the incline. Since there is no friction between 4 kg block and plane below it, the motion of 4 kg block gives us sin …I Since the friction coefficient between the 2 kg block and plane below it is . , the motion of 2 kg block gives sin . ( cos ) … II Rotation of the pulley gives . ( )( . ) ⇒ Substituting values of , √ √ ⇒ (√ ) √

.

in III above √ .

… III

40. A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical

strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings. Three force are acting on the meter stick T1 viz., , where is the mass of the small object. Since the meter stick does not rotate, the torque about center of the meter stick is zero. Thus ( . ) . . . ⇒ . …I . Since the meter stick does not move, its acceleration is zero. Thus . . … II ⇒ . . . .

T2

41. A uniform ladder of length 10.0 m and mass 16 kg is resting against a vertical wall

making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the

10.11

10

Rotational Mechanics ladder by the ground ? What should be the minimum coefficient of friction for the electrician to work safely ? Take force as shown in the figure. Let be the masses of the electrician and the ladder. As the ladder does not slip or rotate, taking torques at ground/ladder contact point, we have ( sin ) ( sin ) ( cos ) ⇒ x 9. x x x 9. x x From the equilibrium of the ladder we have and , , . Thus ( And From equation

N2 37°

mg Mg

N1

f

) we get .

If value of μ is less than this value, the force of friction will not be able to prevent slipping of the ladder.

42. Suppose the friction coefficient between the ground and the ladder of the previous problem is 0.540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder. Let ∴ ⇒ ⇒ ⇒

be the maximum weight of the electrician that ladder can support. Then . . .

. .

(

)

.

.

.

43. A 6.5 m long ladder rests against a vertical wall reaching a height of 6.0 m. A 60 kg man

stands half way up the ladder. (a) Find the torque of force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to man and assuming the wall to be smooth find the force exerted by the ground on the ladder. (a)

Let θ be the angle which the ladder makes with the wall. Then .

cos (b)

)√

(

( . )

.

The torque on the point of contact of ground/ladder is zero as the ladder is in equilibrium. Therefore . cos . sin ⇒

(

√ (

( .

)

.

)

)

N2 θ°

mg

.

N1

f

If F be the force of the ground on the ladder, then

10.12

10

Rotational Mechanics √

44. The door of an almirah is 6 ft high, 1.5 ft wide and weighs 8 kg. The door is supported by

two hinges situated at distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal find this magnitude. We have

. . . . Let the force on the upper hinge A be and on the lower hinge B be with their respective vertical and horizontal components and and and . By equating vertical and horizontal components of the net force on the hinges we get And Also

A D

C

B

read with above two relations gives us

As the door does not rotate through an axis through B and perpendicular to line AB, the torque on hinge B about this axis is zero, we get . . ⇒ Thus √

45. A uniform rod of length L rests against a smooth roller as shown in figure. Find the

friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is θ. Let μ be the required friction coefficient. Let the ground exert normal force N and frictional force on the rod. Since the roller is smooth it exerts no frictional force. Let ’ be the normal force exerted by the roller on the rod. Since the rod does not slip or spin we have by equating the horizontal, vertical forces and the torque through contact point of rod with ground. sin cos cos cos

And

cos ( (

L

N

mg h

θ

⇒ And

N’ cosθ N’ N’ sin θ

f

) sin

cos )

(

) sin

46. A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2

rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation. (b) the speed of the center of the rod and (c) its kinetic energy.

10.13

10

Rotational Mechanics (a)

With usual notation .

( ) ( )

. (b)

( )

.

.

( .

)

.

.

.

(c)

( . )( .

)( )

( . )( .

)( )

.

47. A uniform square plate of mass 2 kg and edge 10 cm rotates about one of its diagonals

under the action of a constant torque of 0.1 N.m. Calculate the angular momentum and the kinetic energy of the plate at the end of the 5th second after the start. The moment of inertia I through the diagonal is equal to moment of inertia through the line passing through center and parallel to the edge, each being equal to half of moment of inertia through line passing center and perpendicular to the plate. ( .

⇒ ⇒

)

. x

. .

⇒ ⇒

.

.

⇒

48. Calculate the ratio of the angular momentum of the earth about its axis due to its

spinning motion to that about the sun due to orbital motion. Radius of the earth = 6400 km and of the orbit of the earth about the sun . km. Let , ω and L be the moment of inertia, angular velocity and angular momentum of the earth about the sun and , the moment of inertia, angular velocity and angular momentum of the earth about its center. Let r and R be the radii of the earth and the orbit of the earth about the sun. Then ( . x ) . x . x

∴

. .

x

.

x

49. Two particles of masses

are joined by a light rigid rod of length . The system rotates at an angular speed ω about an axis through the center of the mass of the system and perpendicular to the rod. Show that the angular momentum of the system is where μ is the reduced mass of the system defined as . Let be the moment of inertia of the system about the center of mass. Then the distances and of the masses and from the center of mass are ( ) ( )

10.14

10

Rotational Mechanics (

⇒

(

) )

⇒

50. A dumb-bell consists of two identical small balls of mass ½ kg each connected to the two

ends of a 50 cm long light rod. The dumb-bell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5.0 N acts on one of the masses in the direction of its velocity for 0.10 s. Find the new angular velocity of the system. The impulse is provided by the force . in the direction of velocity which is perpendicular to the rod as the mass is moving in circular motion. Thus torque of the force is . ( ) . . The moment of inertia of the dumb-bell about line passing through its center and perpendicular to it is ( . )( . ) ⇒ . ( ) If ω be the angular speed of the dumb bell at the end of the impulse, we have

51. A wheel of moment of inertia .

. and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel. Let the suffixes and refer to the initial and final state of the angular motion quantities. Then . . . ( . )( . ) . . . From the conservation of angular momentum we have . ( . ) 9. .

52. A diver having moment of inertia .

about an axis through its center of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease his moment of inertia to . . what will be the new angular speed ? .

Let the suffixes and refer to the initial and final state of the angular motion quantities. Then by conservation of agular momentum we have ( ) .

53. A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per

minute. Two heavy balls form part of the system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from . . . . , what will be the new angular speed ? Let the suffixes and

refer to the initial and final state of the angular motion quantities.

10.15

10

Rotational Mechanics Then by conservation of agular momentum we have (

)

54. A boy is standing on a platform which is free to rotate about its axis. The boy holds an

open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of "the platform plus the boy system" is . . and that of the umbrella is . . . The boy starts spinning the umbrella about the axis at an angular speed of 2 rev/s with respect to himself. Find the angular velocity imparted to the platform. The angular speed of the umbrella is with respect to the boy. Thus if boy rotates with angular speed ω the umbrella rotates with angular speed – ω . But the motion of the platform is due to third law pair torque on it vis-a-vis torque on the umbrella. Thus . . ( ) ⇒ .

55. A wheel of moment of inertia .

. is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel. If I be the moment of inertia of the second wheel, we have from the conservation of angular momentum ) ( ) ( . ( . )( ) ⇒ . .

56. A kid of mass M stands at the edge of a platform of radius R which can be freely rotated

about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event. Let

be the moment of inertia of the platform/boy/ball system. Then ( ) Since the velocity of the ball is tangential to edge of the platform, the angular velocity of the ball is when the boy catches the ball and thus the angular momentum of the ball is ( ) . By conservation of angular momentum this is equal to , the final angular momentum of the system. ⇒ ( )

57. Suppose the platform of the previous problem is brought to rest with the ball in the hand

of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed as seen by his friend. Find the angular velocity with which the platform will start rotating. Since the velocity of the ball is tangential to the edge of the platform, the ball has a linear momentum m when the boy throws it. The angular momentum of the ball when it leaves the boy’s hand is . This imparts a similar angular momentum to the boy/platform system. Again if is the moment of inertia of the platform/boy system,

10.16

10

Rotational Mechanics then And if the angular speed of the platform boy system is ω then

58. Suppose the platform with the kid in the previous problem is rotating in anticlockwise

direction at an angular speed ω. The kid starts walking along the rim with speed relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform. Let be the final angular speed of the platform. Since the kid and the platform are moving in the same direction the angular speed of the kid relative to ground is . By the conservation of the angular momentum we have ( ) ( ) ⇒

59. A uniform rod of mass

and length is struck at an end by a force F perpendicular to the rod for a short time interval . Calculate (a) the speed of the center of the mass, (b) the angular speed of the rod about center of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the center of mass after force has stopped to act. Assume time t is so small that the rod does not appreciably change its direction while the force acts. (a)

Let be the final speed of the center of mass of the rod after the force F acted on one end of the rod for time t. Then ( )

(b)

The torque produced by the force F while it acted on one end of the rod is ⇒

(c)

(

)

Since the rod undergoes both rotation and translation motion, the kinetic energy is given by .

(d)

.

.

.

The angular momentum is given by .

60. A uniform rod of length L lies on a smooth horizontal table. A particle moving on the

table strikes the rod perpendicularly at the end and stops. Find the distance travelled by the center of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic. Let be the velocity of the particle of mass just before the collision and be the linear velocity of the center of the rod and the angular velocity of the rod about its center just after the collision. By the principle of conservation of linear momentum

10.17

10

Rotational Mechanics ⇒ To find

we use the principle of conservation of angular momentum

⇒ If T is the time taken by the rod to turn through a right angle then Since

is constant we have

In case

, the kinetic energy of the particle rod system after the collision is (

(

)

)

(

)

( )

Thus the kinetic energy before and after the collision is conserved. Since the particle stops and separates from the rod after the collision, the linear momentum is also conserved and hence the collision is elastic.

61. Suppose the particle of the previous problem has a mass

and a speed before the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of center of mass C of the system constituting “the rod plus the particle”. (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle about the center of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the center of mass C after the collision. (f) Find the velocity of the center of mass C and the angular velocity of the system about the center of mass after the collision. (a)

Let the velocity of center of mass be after the particle collides with and sticks to the rod. Then by the principle of conservation of linear momentum we have ( ) ⇒

(b)

As particle moves so does the center of mass of the rod particle system. Let the velocity of the particle w.r.t. center of mass, then

(c)

As particle approaches the rod, so does the center of mass C of the rod particle system. Conversely the rod moves w.r.t. the center of mass C. Let be the velocity of the rod w.r.t. C. Then

be

( )

(d)

When the particle is at a distance L/2 from the center of the rod, the distance of the particle from the center of mass of the rod particle system is .

( ) (

)

10.18

10

Rotational Mechanics Using the value of from part (b) the angular momentum of the particle about center of mass C before collision is (

)(

(

)

)

(

)

Again when the particle is at a distance L/2 from the center of the rod, the distance of the center of the rod from the center of mass of the rod particle system is .

( ) (

)

Using the value of from part (c) the angular momentum of the rod about center of mass C before collision is ( (e)

)(

)

(

)

Let I be the moment of inertia of the rod particle system after the collision, then (

* ((

)

(

) (

(f)

)

(

)

) +

(

)

(

)

) (

)

By the principle of conservation of linear momentum we get ( ) ⇒ Again by the principle of conservation of angular momentum we get ( ⇒

(

(

)

(

) (

)

)(

( (

) ( ) )

)

)

(

)

62. Two small balls A and B each of mass , are joined rigidly to the ends of a light rod of

length L. The rod is clamped at the center in such a way that it can rotate freely about the vertical axis through its center. The system is rotated with an angular speed ω about the axis. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find the new angular speed of the rod. The moment of inertia of the rod balls system is ( ) The moment of inertia of the rod, balls, particle system is ( ) If is the new angular velocity, then by the principle of conservation of the angular momentum, we have ⇒

ω

63. Two small balls A and B each of mass , are joined rigidly to the ends of a light rod of length L as in figure. The system translates on a frictionless horizontal surface with a velocity in a direction perpendicular to the rod. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find (a) the linear speeds of the balls A and

10.19

10

Rotational Mechanics B after the collision, (b) the velocity of the system A+B+P and (c) the angular speed of the system about C after the collision. (a)

Since the rod will exert force on ball B only along it length, that is perpendicular to the velocity of B, will not affect the velocity of B. Taking and as the velocities of balls A and B after collision, we get from the principle of conservation of linear momentum

⇒ (b)

(

A

v0

B

v0

L

)

Let be the velocity of the system A+B+P after the collision. Then by the principle of conservation of linear momentum ⇒

(c)

If I be the moment of inertia of the system A+B+P about its center of mass C, then ( ) ( ) The linear momentum of the system before the collision is ( ) The linear momentum of the system after the collision is By the principle of conservation of linear momentum, we have ⇒

64. Suppose the rod with balls A and B of the previous problem is clamped at the center in

such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle of the same mass is dropped from a height on the ball B. The particle collides with B and sticks with it. (a) Find the angular momentum and angular speed of the system just after the collision. (b) What should be the minimum valus of so that the system makes a full rotation after the collision. (a)

Let

be the velocity of the particle when it strikes the ball B. Then √ By the principle of conservation of angular momentum, the angular momentum just after the collision is same as the angular momentum just before the collision. Thus we have √ If

√

be the moment of inertia of the system about the given axis, then ( )

( )

10.20

10

Rotational Mechanics √

⇒ (b)

√

Take the initial horizontal position as the reference position for the potential energy. The system will make a full rotation if the Energy of the system when ball B is at the highest point is greater than zero. But since the center of mass does not move, the energy of the system with ball B at the highest position is given by . ( ) . . ( ) ⇒

65. Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia . . and a radius 2.0 cm. Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant. (a)

The kinetic energy of the system when the 400 g block has descended 50 cm will be equal to the change in the potential energy of the system. . ( ) ⇒ . ( ) .9 m1 m2

(b)

Let the angular velocity of the pulley at the instant when the 400 g block has descended cm be ω. Then speed of the blocks will be . Now ( . .

.

)

.

.

√

⇒

.

.

.

66. The pulley shown in the figure has a radius 20 cm. and moment of inertia .

. . The string going over it is attached at one end to a vertical spring of spring constant 50N/M fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take . Let us take the initial position as the reference position for the potential energy. Let the final position be the position when the 1 kg block has descended . Then by the principle of conservation of energy we get (

) [ ( )(

)( .

)

( .

) ]

.

⇒

.

.

67. A meter stick is held with one end on a rough horizontal floor. It is gently allowed to fall

10.21

10

Rotational Mechanics on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor. If I is the moment of inertia about the end held on floor, then The torque provided by the gravitational force about this axis is ( ) ⇒ ⇒

√

√

( . )

.

√

Aliter: If I is the moment of inertia about the end held on floor, then By the principle of conservation of energy ( ) ⇒

√

√

√

√ 9.

.

68. A meter stick weighing 240 g is pivoted at its upper end in such a way that it can freely

rotate in a vertical plane through this end. A particle of mass 100 g is attached to the upper end of the stick through alight string of length 1 m. Initially the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the sick will rise. Let us denote the mass of the particle by , that of the rod by and the length of the rod by . When the particle falls to the lowest point in its path, its velocity is . If the √ stick rises through a maximum angle θ, then height gained by the lower end where the particle sticks is ( cos ) The height gained by the center of mass of the meter stick is . The moment of inertia of the particle rod system about the axis of rotation is [

( ) ]

Now the angular momentum before and after the collision is conserved because the work done during this interval by the only external force, that is the gravitational force, is negligible. Therefore, if ω is the angular velocity after the collision, we have ⇒

(

)

(

)

Taking the lowest position of the particle as the reference position for the potential energy, we get from the principle of conservation of energy

10.22

10

Rotational Mechanics ⇒

(

)

⇒

cos

⇒

cos

.

(

) ( .

(

)(

)

( .

)(

)

)( .

)

.

.

69. A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle

of 60° and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37° with the vertical. Let be the mass of the rod and , the heights of the lowest tip of the rod when the rod makes angle 60° and 37° respectively. If be the length of the rod, then ( cos ) and ( cos ) Let be the velocity of the mass element and the angular velocity of the rod when it make angle 37° with the vertical. By the principle of conservation of energy we have ( ⇒

√

)

(cos

cos

)

(

)

√ .9

√

⇒ .9 Having found the radial acceleration we now proceed to find the value the tangential acceleration. Since the only force acting on the rod is the gravitational force which acts on the center of mass of the rod. Therefore the torque of the gravitational force on the axis of rotation is given by ( sin ) ( . ) .

∴ ⇒ The force forces ∴

(

)

.9 acting on the mass element and . √ .9√

at the tip of the rod is the resultant of the two

70. A cylinder rolls on a horizontal plane surface. If the speed of the center is 25 m/s, what is the speed of the highest point ?

Under the rolling motion the velocity of the highest point is double the velocity of the center of the rolling body. Therefore the velocity of the highest point is 50 m/s.

71. A sphere of mass

rolls on a plane surface. Find its kinetic energy at an instant when its center moves with speed . The kinetic energy of the sphere is (

)( )

72. A string is wrapped over the edge of a uniform disc and free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.

Let the radius of the disc be . If T be the tension in the string then torque on the disc is

10.23

10

Rotational Mechanics ⇒ Where is the tangential and hence the downward acceleration of the disc. From the downward motion of the disc we have (

)

⇒

73. A small spherical ball is released from a point a height on a rough track shown in figure. Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track. Let the mass of the sphere be and radius . By the principle of conservation of mechanical energy h

.

⇒

√

74. A small disc is rolling with a speed on the horizontal track of the previous problem from right to left. To what height will it climb up the curved part ?

Let the mass of the disc be and radius . If is the height climbed by the disc we have by the principle of conservation of mechanical energy . ⇒

75. A sphere starts rolling down an incline of inclination θ. Find the speed of its center when it has covered length .

When the sphere has rolled down a length , the height scaled down by it is sin By the principle of conservation of mechanical energy we have . Substituting value of √(

A

l

and solving for .

B

sin )

θ

76. A hollow sphere is released from the top of an inclined plane of inclination θ. (a) What

should be the minimum coefficient of friction between the sphere and the plane so as to prevent sliding ? (b) Find the kinetic energy of the ball as it moves down a length on the incline if the friction coefficient is half the value calculated in part (a).

10.24

10

Rotational Mechanics (a)

Let μ be the minimum coefficient of friction between the sphere and the plane required to prevent sliding. From the motion of the hollow sphere, we have sin …I Where is the acceleration along the incline. This is also the tangential acceleration of the rolling sphere. But cos … II And

mg cos θ

mg θ

⇒ Putting in I above and solving for sin From equation II we have tan (b)

When

tan

we have cos

⇒ sin Since the acceleration is constant we have lg sin .

( )

sin

77. A solid sphere of mass

is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reached the bottom of the cup. Let the radius of the hemisphere be R and the mass of the sphere be and radius . By the principle of conservation of mechanical energy . ⇒

√

The normal force consists of the normal reaction force ⇒

.

(

and the centrifugal force )

78. Figure shows a rough track, a portion of which is in the form of a cylinder of radius R.

With what minimum linear speed should a sphere of radius be set rolling on the horizontal part so that it completely goes round the circle on cylindrical part. Let

be the minimum starting linear speed required for the sphere to go round the circle

10.25

10

Rotational Mechanics on the cylindrical path. Let the meeting point of the horizontal path with the cylindrical portion be the reference point for the potential energy. Let , refer to the initial state and the state when the sphere is at the top of the cylindrical path. Then by the principle of conservation of mechanical energy ( )

R

For to be minimum the normal force on the sphere at final position should be zero. Therefore ( ⇒

)

(

) (

⇒ ⇒

√

(

)

(

)

(

)

(

)

)

79. Figure shows a spherical ball of mass

rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R. (a) Find the kinetic energy of the ball when it is at a point A where the radius makes an angle θ with the horizontal. (b) Find the radial and tangential accelerations of the center when ball is at A. (c) Find the normal force and the frictional force acting on the ball if H = 60 cm, R=10 cm, θ = 0 and . (a) R H

A

θ

Take the lowest point of the track as the reference point for the potential energy. Let , refer to the positions when the ball is at the lowest point and at the point A respectively. Then by the principle of conservation of mechanical energy ( sin ) ( ) ⇒ ⇒ sin ) (b)

(

(

sin )

)

(

From part (a) above we have ( sin ) ⇒ ( sin ) To find the tangential acceleration differentiate the above first equation

10.26

10

Rotational Mechanics (

cos ⇒ (c)

)

cos

cos

Using values of

and .

from part (b) above (

(

.

)) cos

( .

(

. .

) )

.9 x9. ( )

. 9

80. A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally with a cue. Where should it be hit so that the shell does not slip on the surface ? If it is hit at the center of the shell, the force shall produce no torque and instead of rolling it will slip on the surface. If hit below that it will get projected at the an angle above the horizontal. So it has to be hit above the center. Suppose it is hit at a distance above the center. Since the shell does not slip on the surface we have Since the gravitational force acts on the shell at its center, it produces no torque and therefore the angular momentum, about a horizontal axis through the center of the shell, is conserved. By the principle of conservation of angular momentum we have ⇒

81. A uniform wheel of radius R is set into rotation about its axis at an angular speed ω. This

rotating wheel is now placed on a rough horizontal surface with its axis horizontal. Because of the friction at the contact, the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling on the surface. Find the linear speed of the wheel after it starts pure rolling. By the principle of conservation of angular momentum Where the left side is the initial angular momentum and on the right side the angular momentum after the pure rolling has started. ⇒ ( ) ⇒

(

)

82. A thin spherical shell lying on a rough horizontal surface is hit sharply and horizontally

by a cue in such a way that the line of action passes through the center of the shell. As a result the shell starts moving with linear speed without any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface. Because of the friction with the surface of contact the shell will start rotation and will eventually start pure rolling. Let be the final linear velocity and the final angular velocity when pure rolling starts. Since the gravitational force acts on the shell at its center, it produces no torque and therefore the angular momentum, about a horizontal

10.27

10

Rotational Mechanics axis through the center of the shell, is conserved. By the principle of conservation of angular momentum we have ( ) ⇒

83. A hollow sphere of radius R lies on a smooth horizontal surface. It is pulled by a

horizontal force acting tangentially from the highest point. Find the distance travelled by the sphere during the time it makes one full rotation. ⇒ Since α is constant we have ⇒ ⇒ But ⇒

.

.

.

.

(

)

84. A solid sphere of mass 0.50 kg is kept on a horizontal surface. The coefficient of static

friction between the surface in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface ? Let force be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface. Since the sphere is rolling the frictional force is in the same direction as the force applied at its highest point. Then ( ) ( ) ⇒ ... I and ⇒ … II From I and II above (

⇒

(

)

)

⇒ The force ⇒

will be maximum when the friction coefficient has its maximum value . . .

.

85. A solid sphere is set into motion on a rough horizontal surface with linear speed in the

forward direction and an angular speed in the anticlockwise direction as shown in the figure. Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts. (a)

Since the linear velocity is in the forward direction and the angular velocity in

10.28

10

Rotational Mechanics anti-clock wise direction, the sphere necessarily slides. Let the mass of the sphere be and the linear velocity when rotation stops be . We get from the principle of conservation of angular momentum about the point of contact ( ) ⇒ (b)

Let and be the linear and angular speeds of the sphere when pure starts rolling. Then by the principle of conservation of angular momentum ( ) ⇒

.

⇒

86. A solid sphere rolling on a rough horizontal surface with a linear speed collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in backward direction.

Since the sphere collides elastically, it will bounce back with same speed linear but angular speed continues to be as it was before collision. Let and be the linear and angular speeds of the sphere when pure starts rolling. By the principle of conservation of angular momentum at the initial state of backward motion and when pure rolling starts ( ) ⇒

. ∎

10.29