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A ring rolls down an inclined plane. The ratio of the rotational kinetic energy to translational kinetic energy is
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Verified Answer
The correct answer is:
$1: 1$
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, $K_{T}=\frac{1}{2} M v^{2}$
$$
\therefore K_{R}=K_{T} \text { or } K_{R}: K_{T}=1: 1
$$
$$
K_{R}=\frac{1}{2} I \omega^{2}
$$
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, $K_{T}=\frac{1}{2} M v^{2}$
$$
\therefore K_{R}=K_{T} \text { or } K_{R}: K_{T}=1: 1
$$
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