Question

If a sphere is rolling, then the ratio of its rotational kinetic energy to the total kinetic energy is

• A. 1 : 2
• B. 2 : 5
• C. 2 : 7
• D. 5 : 7

Answer 1

Answer: (C)

Let m be the mass, r the radius of the sphere and let v and $\omega$ be the linear and angular velocities in rolling down.
Thus total kinetic energy = linear kinetic energy + rotational kinetic energy
$=\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2$
where I is the moment of inertia ie, $I=\frac{2}{5}m{r}^2$
Hence, total kinetic energy$=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)\frac{v^2}{r^2}$
$\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2$
$\frac{7}{10}m{v}^2$
So, the ratio$=\frac{RotationalKE}{TotalKE}$
=$\frac{\frac{1}{5}m{v}^2}{\frac{7}{10}m{v}^2}=2:7$