Question

A ring rolls down an inclined plane. The ratio of the rotational kinetic energy to translational kinetic energy is:

• A. $1:3$
• B. $1:1$
• C. $3:1$
• D. $2:1$

Answer 1

The correct option is B $1:1$

For a rolling ring:

Rotational Kinetic energy $K_R=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(m{R}^2\right){\left(\frac{v}{R}\right)}^2=\frac{1}{2}m{v}^2=K_T$
$\Rightarrow \frac{K_R}{K_T}=1$