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MHT CET 2024 (4 May, Shift 2) · Rolling Motion

Rolling Motion question from MHT CET 2024 (4 May, Shift 2)

Q25MHT CET 2024 · 4 MAY · SHIFT 2Easy

Rolling Motion

A solid cylinder of mass ' $M$ ' and radius ' $R$ ' rolls down an inclined plane of height ' $h$ '. When it reaches the foot of the plane, its rotational kinetic energy is ( $g =$ acceleration due to gravity)

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Answer: (a)

Total KE at the foot $= Mgh$ .
Solid cylinder: $\frac{K_{rot}}{K_{total}} = \frac{1/2}{3/2} = \frac{1}{3}$
$K_{rot} = \frac{Mgh}{3}$
So the answer is (a) $\frac{Mgh}{3}$ .

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