Question

A spherical ball rolls on an inclined plane without slipping. Ratio of total kinetic energy to rotational kinetic energy is

• A. $\frac{2}{7}$
• B. $\frac{2}{5}$
• C. $\frac{7}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is C $\frac{7}{2}$

According to question -


Spherical ball is rotating about axis passing through it's centre of mass with angular speed $\omega$ and COM is translating with speed $v$.
Now,
Total Kinetic Energy $=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$
$=\frac{1}{2}\times \frac{2}{5}m{R}^2\times \frac{v^2}{R^2}+\frac{1}{2}m{v}^2$
$[\because$ $I=\frac{2}{5}m{R}^2$ for sphere, $v=\omega R$ for pure rolling $]$
$=\frac{7}{10}m{v}^2$ ............(1)
Rotational Kinetic Energy $=\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\times \frac{2}{5}m{R}^2\times \frac{v^2}{R^2}$
$=\frac{m{v}^2}{5}$ ...........(2)
On dividing equation (1) by (2),
$\frac{\text{Total K.E}}{\text{Rotational K.E}}=\frac{7}{2}$
Ratio of Total K.E to Rotational K.E is $\frac{7}{2}$.