Question

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

• A. 1:03
• B. 1:01
• C. 3:01
• D. 2:01

Answer 1

Answer: (B)

1:01

Answer 2

Translational kinetic energy,
$K_T = \frac{1}{2} mv^2$
where $m$ = mass of the ring,
$v$ = speed of the centre of mass of the ring
Rotational kinetic energy,
$K_R = \frac{1}{2} I \omega^2$
Here, $I = mR^2 , \omega = \frac{v}{R}$
So, $K_R = \frac{1}{2} ( mR^2 ) \left(\frac{v}{R} \right)^2 = \frac{1}{2} mv^2$
$\therefore\:\:\:\: \frac{K_R}{K_T} = \frac{1}{1}$