Question

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 p rad s-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is μ _k = 0.2.

Answer 1

Given, the radius of both the solid disc and the ring is 10 cm and their initial angular velocity is 10π  rad-s −1 . The coefficient of kinetic friction is 0.2.

The expression for the linear velocity at the point of contact is,

v 0 =R ω 0

Here, R is the radius of the circular object and ω 0 is the initial angular speed.

Substituting the given values in the above equation, we get:

v 0 =( 10× 10 −2 )( 10π ) =π m/s

The expression for the normal force acting at the point of contact is,

N=mg

Here, m is the mass of the circular object and g is the gravitational acceleration.

The expression for the frictional force acting at the point of contact is,

f k = μ k N

Here, μ k is the coefficient of kinetic friction and N is the normal force acting at the point of contact.

Substituting mg for N in the above equation, we get:

f k = μ k mg

The net force acting at the point of contact is,

F net =ma

Here, a is the net acceleration.

The only force acting at the point of contact is the frictional force. Therefore,

F net = f k

Substituting ma for F net and μ k mg for f k in the above equation, we get:

ma= μ k mg a= μ k g

The expression for linear acceleration at the point of contact is,

a=Rα

The kinematic equation for the motion at the point of contact is,

v= v 0 +at

Here, v is the final velocity and v 0 is the initial velocity at the point of contact and t is the time taken to start the rolling motion.

Substituting the values in the above equation, we get:

0=π+( Rα )t t= −π Rα …… (1)

The expression for net torque acting on the circular object is,

f k ⋅R=Iα

Solving for α, the equation is rearranged as,

α= f k R I

Substituting f k R I for α in the equation (1), we get:

t= −π R( f k R I ) =− πI f k R 2

Here, kinetic frictional force f k and radius R of both solid disc and ring are the same. Therefore,

t∝I

Moment of inertia of the solid disc is,

I D = 1 2 M R 2

Moment of inertia of the ring is,

I R =M R 2

The moment of inertia of the solid disc is less than that of the ring,

I D < I R

The time takento start the rolling motion is directly proportional to the moment of inertia of the object. Therefore,

t D < t R

Thus, the solid disc starts the rolling motion earlier than the ring.