MHT CET · Physics · Rotational Motion
The rotational kinetic energy and translational kinetic energy of a rolling body are same, the body is
- A disc
- B sphere
- C cylinder
- D ring
Answer & Solution
Correct Answer
(D) ring
Step-by-step Solution
Detailed explanation
Translational kinetic energy of a ring is:
\(\mathrm{KE}_{\text {Trans }}=\frac{1}{2} \mathrm{mv}^2\)
Rotational kinetic energy of the ring is:
\(\begin{aligned}\mathrm{KE}_{\text {Rolling }} & =\frac{1}{2} \mathrm{I} \omega^2 \\& =\frac{1}{2} \mathrm{mR}^2 \omega^2 \\
\mathrm{KE}_{\text {Rolling }} & =\frac{1}{2} \mathrm{mv}^2 \quad(\because \mathrm{v}=\omega \mathrm{R})\end{aligned}\)
\(\mathrm{KE}_{\text {Trans }}=\frac{1}{2} \mathrm{mv}^2\)
Rotational kinetic energy of the ring is:
\(\begin{aligned}\mathrm{KE}_{\text {Rolling }} & =\frac{1}{2} \mathrm{I} \omega^2 \\& =\frac{1}{2} \mathrm{mR}^2 \omega^2 \\
\mathrm{KE}_{\text {Rolling }} & =\frac{1}{2} \mathrm{mv}^2 \quad(\because \mathrm{v}=\omega \mathrm{R})\end{aligned}\)
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