Question

Two identical hollow spheres having mass $m=3\text{kg}$ and radius $r=1\text{m}$ are performing combined translational and rotational motion as shown in the figure. If $\omega =2\text{rad/s}$,$v_{com}=3\text{m/s}$ and ${\omega }^{\prime }=3\text{rad/s},v_{com}^{{}^{\prime }}=2\text{m/s}$, find the ratio of their kinetic energies.

• A. $\frac{3}{7}$
• B. $\frac{7}{6}$
• C. $\frac{8}{7}$
• D. $\frac{3}{5}$

Answer 1

The correct option is B $\frac{7}{6}$

Given, $m=3\text{kg},r=1\text{m},\omega =2\text{rad/s},v_{com}=3\text{m/s},{\omega }^{\prime }=3\text{rad/s},v_{com}^{{}^{\prime }}=2\text{m/s}$
Moment of inertia of hollow sphere;$I=\frac{2}{3}m{r}^2=\frac{2}{3}\times 3\times 1^2=2{\text{kg-m}}^2$.
Since, both sphere are identical therefore, $I$ will be same.

We know that,
Total kinetic energy for combined translational and rotational motion is
$=\frac{1}{2}m{v}_{\text{COM}}^2+\frac{1}{2}I{\omega }^2$
$\Rightarrow K.E=\frac{1}{2}m{v}_{\text{COM}}^2+\frac{1}{2}I{\omega }^2$
So, for first sphere,
$K.E=\frac{1}{2}\times 3\times 3^2+\frac{1}{2}\times 2\times 2^2=17.5\text{J}$
Similarly, for other sphere
$K.E^{{}^{\prime }}=\frac{1}{2}m{v}_{com}^{\prime 2}+\frac{1}{2}I{\omega }^{\prime 2}$
$=\frac{1}{2}\times 3\times 2^2+\frac{1}{2}\times 2\times 3^2=15\text{J}$

$\therefore \frac{K.E}{K.E^{{}^{\prime }}}=\frac{17.5}{15}=\frac{7}{6}$