MHT CET · Physics · Rotational Motion
A solid cylinder of radius 'R' and mass 'M' rolls down an inclined plane of height 'h'. When it reaches the bottom of the plane, its rotational kinetic energy is \((\mathrm{g}=\) acceleration due to gravity)
- A \(\frac{\text { Mgh }}{3}\)
- B Mgh
- C \(\frac{\text { Mgh }}{2}\)
- D \(\frac{\mathrm{Mgh}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\text { Mgh }}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \text { Total } \mathrm{KE} &=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} \\ &=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{1}{2} \mathrm{MR}^{2}\right) \mathrm{v}^{2} \\ &=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{Mv}^{2}=\frac{3}{4} \mathrm{Mv}^{2}=\mathrm{Mgh} \end{aligned}\)
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