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

$\frac{K^{2}}{K^{2} + R^{2}}$

$\frac{R^{2}}{K^{2} + R^{2}}$

$\frac{K^{2} + R^{2}}{R^{2}}$

$\frac{K^{2}}{R^{2}}$

Solution

$\frac{K^{2}}{K^{2} + R^{2}}$

Kinetic energy of rotation is

K_rot = $\frac{1}{2} {Iω}^{2}$

= $\frac{1}{2} {MK}^{2} \frac{v^{2}}{R^{2}}$

where, k is radius of gyration.

Kinetic energy of translation is K_tran = $\frac{1}{2} {Mv}^{2}$

Thus, total energy,

E = K_rot + K_trans

= $\frac{1}{2} {MK}^{2} \frac{v^{2}}{R^{2}} + \frac{1}{2} {Mv}^{2}$

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

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

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

Hence

$\frac{K_{rot}}{Total energy, E} = \frac{\frac{1}{2} {MK}^{2} \frac{v^{2}}{R^{2}}}{\frac{1}{2} \frac{M v^{2}}{R^{2}} (K^{2} + R^{2})}$

$\frac{K_{rot}}{Total energy E} = \frac{K^{2}}{K^{2} + R^{2}}$

Some More Questions From System of Particles and Rotational Motion Chapter

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