Q143 / 180 · COMEDK 2012 · Physics
A ring rolls down an inclined plane. The ratio of the rotational kinetic energy to translational kinetic energy is
- A$1: 3$
- Correct answerB$1: 1$
- C$3: 1$
- D$2: 1$
Solution
The rotational kinetic energy of a ring is
$$
K_{R}=\frac{1}{2} I \omega^{2}
$$
For a ring, $\quad {I}=M R^{2}$ and $\omega=\frac{v}{R}$
$$
\therefore \quad K_{R}=\frac{1}{2}\left(M R^{2}\right) \times\left(\frac{v}{R}\right)^{2}=\frac{1}{2} M v^{2}
$$
Also, translational kinetic energy, $K_{T}=\frac{1}{2} M v^{2}$
$$
\therefore K_{R}=K_{T} \text { or } K_{R}: K_{T}=1: 1
$$
Hence, the correct answer is option: 'B'.