Question

Two plates $1$ and $ 2$ move with velocities $-v$ and $2\, v$ respectively. If the sphere does not slide relative to the plates, assuming the masses of each body as $m$. If $m v^{2}=\frac{120}{123}$, then the kinetic energy of the system (plates $+$ sphere) is ___.

Answer 1

Top point of sphere in having velocity $v(\leftarrow)$ and bottom point in moving with $2 v(\rightarrow)$
We can find position of IRC
$\frac{y}{v}=\frac{2 R-y}{2 v} $
$\Rightarrow y=\frac{2}{3} R$
Hence we can assure the sphere will have pure rotation about $C(ICR)$
Hence the motion of sphere can be considered as pure rotation about $C$.

The angular velocity of sphere $\omega=\frac{v}{y}=\frac{3 v}{2 R}$
The kinetic energy of sphere $k_{\text {rotation }}=\frac{1}{2} I_{c} \omega^{2}$
$\Rightarrow k_{\text {rotationa }}=\frac{1}{2}\left[\frac{2}{5} m R^{2}+m( OC )^{2}\right] \omega^{2}$
$O C=R-\frac{2}{3} R=\frac{R}{3}$
$K_{\text {rotational }}=\frac{1}{2}\left(\frac{23}{45} m R^{2}\right) \omega^{2}=\frac{23}{90} m \omega^{2} R^{2}$
$=\frac{23}{90} m\left(\frac{3 v}{2 R}\right)^{2} R^{2}=\frac{23}{40} m v^{2}$
Hence total kinetic energy
$K_{\text {total }}=\frac{1}{2} m v^{2}+\frac{1}{2} m(2 v)^{2}+\frac{23}{40} m v^{2}=\frac{123}{40} m v^{2}$