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MHT CET 2020 (16 Oct, Shift 1) · Rolling Motion

Rolling Motion question from MHT CET 2020 (16 Oct, Shift 1)

Q63MHT CET 2020 · 16 OCT · SHIFT 1Easy

Rolling 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, then its rotational kinetic energy is ( $g =$ acceleration due to gravity)

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Answer: (d)

Total KE at the bottom $= Mgh$ .
Solid cylinder: $K_{rot} = \frac{1}{4}Mv^2$ , $K_{total} = \frac{3}{4}Mv^2$ , so $K_{rot} = \frac{1}{3}K_{total}$
$K_{rot} = \frac{Mgh}{3}$
So the answer is (d) $\frac{Mgh}{3}$ .

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