System Of Particles And Rotational Motion Ques 6
6. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $\mathrm{K}$. If radius of the ball be $\mathrm{R}$, then the fraction of total energy associated with its rotational energy will be
[2003]
(a) $\frac{R^2}{K^2+R^2}$
(b) $\frac{K^2+R^2}{R^2}$
(c) $\frac{K^2}{R^2}$
(d) $\frac{K^2}{K^2+R^2}$
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Answer:
Correct Answer: 6.(d)
Solution: (d) As we know that,
Rotational energy $=\frac{1}{2} I(\omega)^2=\frac{1}{2}\left(m K^2\right) \omega^2$ And,
Linear kinetic energy $=\frac{1}{2} m \omega^2 R^2$
$\therefore$ Required fraction,
$ =\frac{\frac{1}{2}\left(m K^2\right) \omega^2}{\frac{1}{2}\left(m K^2\right) \omega^2+\frac{1}{2} m \omega^2 R^2}=\left(\frac{K^2}{K^2+R^2}\right) $
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