Question

Two beads each of mass $m$ are fixed on a light rigid rod of length $2l$ which is free to rotate in a horizontal plane. The bead on the far end is given some velocity $v$ as shown in the figure. If $K_{cm}$ represents the kinetic energy of the centre of mass of the system and $K_R$ represents the rotational kinetic energy of the system, then what is the value of $\frac{K_{cm}}{K_R}$ ?

Answer 1

Answer: 0.9

The distance of the centre of mass from the hinge,
$r_{cm}=\frac{ml+m(2l)}{m+m}=\frac{3}{2}l$
The angular velocity of the system,
$\omega =\frac{v}{2l}$
The velocity of the centre of mass of the system,
$v_{cm}=\frac{3l}{2}\omega =\frac{3}{4}v$
$\Rightarrow K_{cm}=\frac{1}{2}2m{v}_{cm}^2=\frac{9m{v}^2}{16}$
The moment of inertia of the system about the hinge,
$I=m{l}^2+m{\left(2l\right)}^2=5m{l}^2$
The rotational kinetic energy of the system,
$K_{\mathrm{R}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}5m{l}^2{\left(\frac{v}{2l}\right)}^2$
$\Rightarrow K_R=\frac{5m{v}^2}{8}$
$\frac{K_{cm}}{K_R}=\frac{9m{v}^2}{16}\times \frac{8}{5m{v}^2}=0.9$