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Q.
A coin of mass $ 10\,g $ rolls along a horizontal table with a velocity of $ 6\,cm/s $ . Its total kinetic energy is
J & K CETJ & K CET 2012System of Particles and Rotational Motion
Solution:
The total kinetic energy of the coin is
$K=K_{\tau}+K_{R} $
$=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
For coin, the moment of inertia
$I=\frac{1}{2} M R^{2} $
$\therefore K=\frac{1}{2} M v^{2}+\frac{1}{4} M v^{2} $
$(\because v=r \omega) $
$=\frac{3}{4} M v^{2} $
$=\frac{3}{4} \times\left(10 \times 10^{-3}\right)\left(6 \times 10^{-2}\right)^{2} $
$=27 \times 10^{-6} J=27 \,\mu \,J$