Q31 of 33 Page 178

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 300. The coefficient of static friction μs = 0.25.

A. How much is the force of friction acting on the cylinder?


B. What is the work done against friction during rolling?


C. If the inclination q of the plane is increased, at what value of q does the cylinder begin to skid, and not roll perfectly?

Given,


Mass of the cylinder, m = 10 kg


Radius of the cylinder, r = 15 cm = 0.15 m


Inclination angle, θ = 300


Coefficient of static friction, μs = 0.25


The moment of inertia of the cylinder about its geometric axis is given by, I = 0.5mr2


The free diagram of the body is given by,


31


We have,


A. From the newton second law of motion, force net,


Fnet = ma


Where,


m = mass of the body


a = acceleration of the body


∴ mg sinθ – f = ma


⇒ f = ma – mg sinθ


= (10×3.27)-(10×9.81×sin30)


= 16.3 N


B. During rolling the instantaneous point of contact have zero velocity. Thus, work done against frictional force is zero.


C. For rolling without skid, we have,


⇒ 3μ = tan θ


More from this chapter

All 33 →
29

Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

A. Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.


B. What is the force of friction after perfect rolling begins?

 30

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad/s. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk= 0.2.

 32

Read each statement below carefully, and state, with reasons, if it is true or false;

A. During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.


B. The instantaneous speed of the point of contact during rolling is zero.


C. The instantaneous acceleration of the point of contact during rolling is zero.


D. For perfect rolling motion, work done against friction is zero.


E. A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

 33

Separation of Motion of a system of particles into motion of the center of mass and motion about the center of mass:

A. Show


Where pi is the momentum of the ith particle (of mass mi) and Note is the velocity of the ith particle relative to the canter of mass.


Also, prove using the definition of the centre of mass


B. Show


Where K is the total kinetic energy of the system of particles, K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV2/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.


C. Show


Where is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember; rest of the notation is the standard notation used in the chapter. Note and MR × V can be said to be angular momenta, respectively, about and of the canter of mass of the system of particles.


D. Show


Further, show that


Where is the sum of all external torques acting on the system about the canter of mass.


(Hint: Use the definition of canter of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)