Question

A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Answer 1

Given, the mass of the hoop is 100 kg, radius of the hoop is 2 m and the initial velocity of the centre of mass is 20 cm/s.

Let m be the mass of the hoop and r be the radius of the hoop.

The equation for the moment of inertia of the hoop is,

I=m r 2

The equation for the initial velocity of the centre of mass of the hoop is,

v cm =r ω 0

Here, ω 0 is the initial angular velocity of the hoop.

The equation for the initial kinetic energy of the hoop is

K 0 = 1 2 m v cm 2 + 1 2 I ω 0 2

Substitute the given values in the above equation.

K 0 = 1 2 m v cm 2 + 1 2 ( m r 2 ) ω 0 2 = 1 2 m v cm 2 + 1 2 m ( r ω 0 ) 2 = 1 2 m v cm 2 + 1 2 m ( v cm ) 2 =m v cm 2

Substitute the given values in the above equation.

K 0 =( 100 kg ) ( 20  cm/s ) 2 =( 100 kg ) { ( 20  cm/s )( 1 m 100 cm ) } 2 =( 100 kg ) ( 0.20 m/s ) 2 =4 J

Finally, the hoop comes to rest. Therefore, the final kinetic energy of the hoop is zero, that is, K f =0 .

The equation for the change in the kinetic energy of the hoop system is,

| ΔK |=| K f − K 0 |

Here, ΔK is the change in the kinetic energy of the mass system.

Substitute the values in the above equation.

| ΔK |=| 0−4 J | =4 J

The expression for work energy theorem is,

W=ΔK

Here, W is the amount of work done on the system.

Substitute the calculated values in the above equation.

W=4 J

Thus, the amount of work done to stop the rolling motion of the hoop is 4 J.