Question

The speed of rolling of a ring of mass m changes from v to 3v. what is the change in kinetic energy. then the percentage increase in its angular momentum will be

Answer 1

Let , m = mass of the ring

r= radius of the ring

v = initial velocity

Kinetic energy = translational K.E + rotational K.E

= 1/2 mv² + 1/2 I w²

I is the moment of inertia , I= mr²

w = v/r

K.E initial = 1/2 mv²+ 1/2 mr² × v²/r²

=> mv²

v' = 3v = final velocity

K.E final = 1/2 mv'² + 1/2 Iw'²

=> 3/2 mv² + 1/2 mr² × 9v²/r²

=> 6 mv²

Change in Kinetic Energy = k.E final - KE initial

=> 6mv² - mv²

=> 5mv²

For Angular momentum -

Initial Angular momentum= Iw

=> mr²× v/r

=> mvr

Final Angular momentum = mr² × 3v/r

=> 3 mvr

Change in Angular momentum = 3mvr - mvr

= 2mvr

% change in Angular momentum

= 2mvr/mvr × 100

= 200%