System of Particles and Rotational Motion - Result Question 81
88. A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is
[1994]
(a) $2 / 5$
(b) $2 / 7$
(c) $3 / 5$
(d) $3 / 7$
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Answer:
Correct Answer: 88. (b)
Solution:
(b) $\frac{K_r}{E}=\frac{\frac{1}{2} M K^{2} \omega^{2}}{\frac{1}{2} M \omega^{2}[K^{2}+R^{2}]}=\frac{K^{2}}{K^{2}+R^{2}}$
$ =\frac{2 / 5}{1+2 / 5}=\frac{2}{7} $
Here, $K^{2}=\frac{2}{5} R^{2}$
When a body is executing only linear motion, its K.E. is given by $E_t=\frac{1}{2} m v^{2}$ when a body is rolling without slipping, its centre of mas has linear motion too.
Therefore, its total kinetic energy
$E=E_t+E_r$
$=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} m(K^{2}) \frac{v^{2}}{R^{2}}$
$=\frac{1}{2} m v^{2}(1+\frac{K^{2}}{R^{2}})$
