Question

The ratio of rotational kinetic energy and translatory kinetic energy of a rolling circular disc is:

• A. $4:1$
• B. $2:1$
• C. $1:4$
• D. $1:2$

Answer 1

Answer: (D)

$1:2$

Rotational kinetic energy = $\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$

v = R $\times \omega$
Translational K.E=$\frac{1}{2}m{v}^2$
$\Rightarrow ratio=\frac{\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)}{\frac{1}{2}m{v}^2}=\frac{1}{2}$