Question

Two plates $1$ and $2$ move with velocities $-v$ and $2v$ 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

Answer: 3

Top point of sphere in having velocity $v(\leftarrow )$ and bottom point in moving with $2v(\to )$
We can find position of IRC
$\frac{y}{v}=\frac{2R-y}{2v}$
$\Rightarrow y=\frac{2}{3}R$
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 $\omega =\frac{v}{y}=\frac{3v}{2R}$
The kinetic energy of sphere $k_{\text{rotation }}=\frac{1}{2}{I}_c{\omega }^2$
$\Rightarrow k_{\text{rotationa }}=\frac{1}{2}\left[\frac{2}{5}m{R}^2+m(OC{)}^2\right]{\omega }^2$
$OC=R-\frac{2}{3}R=\frac{R}{3}$
$K_{\text{rotational }}=\frac{1}{2}\left(\frac{23}{45}m{R}^2\right){\omega }^2=\frac{23}{90}m{\omega }^2{R}^2$
$=\frac{23}{90}m{\left(\frac{3v}{2R}\right)}^2{R}^2=\frac{23}{40}m{v}^2$
Hence total kinetic energy
$K_{\text{total }}=\frac{1}{2}m{v}^2+\frac{1}{2}m(2v{)}^2+\frac{23}{40}m{v}^2=\frac{123}{40}m{v}^2$