A circular disc reaches from top to bottom of an inclined plane of length l. When it slips down the plane, it takes t s. When it rolls down the plane then it takes ((α)/(2))1/2 t s, where α is _________. Numerical answer type. Work it out, then open the answer.

Answer: 3. Slipping (no friction): a = gsin θ, t = √((2l)/(gsin θ)). Rolling disc: a' = 2/3gsin θ, t' = √((3l)/(gsin θ)) = √(3/2) t. Comparing with ((α)/(2))1/2t: α = 3.

A circular disc reaches from top to bottom of an inclined plane of… | JEE Main 2024 (9 Apr, Shift 2)

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Rolling Motion

A circular disc reaches from top to bottom of an inclined plane of length $l$ . When it slips down the plane, it takes $t\text{ s}$ . When it rolls down the plane then it takes $\left(\frac{\alpha}{2}\right)^{1/2} t\text{ s}$ , where $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Slipping (no friction): $a = g\sin\theta$ , $t = \sqrt{\dfrac{2l}{g\sin\theta}}$ .
Rolling disc: $a' = \frac{2}{3}g\sin\theta$ , $t' = \sqrt{\dfrac{3l}{g\sin\theta}} = \sqrt{\dfrac{3}{2}}\,t$ .
Comparing with $\left(\frac{\alpha}{2}\right)^{1/2}t$ : $\alpha = 3$ .

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