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 the ball be R, then the fraction of total energy associated with its rotational energy will be:

• A. \(\frac {K^2+R^2}{R^2}\)
• B. \(\frac {K^2}{R^2}\)
• C. \(\frac {K^2}{K^2+R^2}\)
• D. \(\frac {R^2}{K^2+R^2}\)

Answer 1

The Correct Option is C

Moment of Inertia \(I =mK^2\)
We know that, 
\(v=Rω\)

\(ω = \frac vR\)

Translational Kinetic Energy = \(\frac 12mv^2\)

Rotational Kinetic Energy  = \(\frac 12Iω^2 \)= \(\frac {mK^2v^2}{2R^2}\)

Total Energy = Translational Kinetic Energy + Rotational Kinetic Energy

Total Energy = \(\frac 12mv^2\)+ \(\frac {mK^2v^2}{2R^2}\)

Total Energy = \(\frac 12mv^2\)\((1+\frac {K^2}{R^2})\)

Required fraction = \(\frac {\text {Rotational\  Kinetic\  Energy}}{ \text {Total\  Energy }}\)

= \(\frac {\frac {mK^2v^2}{2R^2}}{ \frac12mv^2 (1+\frac {K^2}{R^2})}\)

= \(\frac {\frac {K^2}{R^2} }{ 1+\frac {K^2}{R^2}}\)

= \(\frac {K^2}{K^2+R^2}\)

So, the correct option is (C): \(\frac {K^2}{K^2+R^2}\)