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6.System of Particles and Rotational Motion
medium
A solid sphere of mass $500\ gm$ and radius $10\ cm$ rolls without slipping with the velocity $20\ cm/s$. The total kinetic energy of the sphere will be ........ $J$
A
$0.014$
B
$0.028$
C
$280$
D
$140$
Solution
since it is pure rolling, $V=\omega R$
$\Rightarrow \omega=\frac{V}{R}=\frac{0.2}{0.1}=2 \mathrm{rad} / \mathrm{s}$
$K E_{\text {Translational}}=\frac{1}{2} m v^{2}=\frac{1}{2} \times \frac{1}{2} \times(0.2)^{2}$
$=10^{-2} J$
$K E_{\text {Rotational}}=\frac{1}{2} J \omega^{2}=\frac{1}{2}\left(\frac{2}{5} M R^{2}\right) \omega^{2}$
$=\frac{\left(\frac{1}{2}\right)(0.1)^{2}}{5}(2)^{2}=4 \times 10^{-3} J$
$K E_{T o t a l}=K E_{T r a n s l a t i o n a l}+K E_{r o a t a t i o n a l}$
$=10^{-2} J+4 \times 10^{-3} J$
$=0.014 J$
Standard 11
Physics
