Question

A ball rolls without slipping .The radius of gyration of the ball about an axis passing through its centre of mass is K .If radius of ball is R,then the fraction of total energy associated with its rotational energy will be

Answer 1

moment of inertia = I= MK²

1. ball is a solid sphere
I= 2/5 MR²
 rotational KE= 1/2w²= 1/2 MK²w²

total energy= 1/2 Iw² +1/2 Mv²

rolling, without slipping
v= Rw
total energy= 1/2 Iw² +1/2 MR²v²
1/2 MK²w² + 1/2 MR²w²
w²M/2= (K² + R²)

Fraction of total Energy associated with KE
1/2MK²w² divided by w²M/2 (K²+R²) = K²/ K²+ R²

Answer 2

Here,

Moment of Inertia I= MK2

I=MK2

Case 1. The ball is a solid sphereI=25MR2

Now rotational KE = 12Iω2=12MK2ω2

Total energy = 12Iω2+12Mv2

With the rolling without sliping condition,

we havev=RωTotal energy = 12Iω2+12MR2ω2=12MK2ω2+12MR2ω2=ω2M2(K2+R2)

Now the fraction of total energy associated with its KE will be

12MK2ω2ω2M2(K2+R2)=K2K2+R2