A circular disc reaches from top to bottom of an inclined plane of length . When it slips down the plane, if takes s. When it rolls down the plane then it takes , where is

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Solution

Q.
A circular disc reaches from top to bottom of an inclined plane of length $l$ . When it slips down the plane, if takes $t$ s. When it rolls down the plane then it takes ${(\frac{α}{2})}^{1 / 2}$ $t$ $s$ , where $α$ is _______ . [2024]

Ans.
(3) If disc slips on inclined plane, then it's acceleration, $a_{1} = g \sin θ$

$l = \frac{1}{2} a_{1} t_{1}^{2} \Rightarrow t = \sqrt{\frac{2 l}{g \sin θ}}$ $\dots (i)$

  If disc rolls on inclined plane its acceleration,

   $a_{2} = \frac{g \sin θ}{1 + \frac{I}{m R^{2}}} = \frac{g \sin θ}{1 + \frac{m R^{2}}{2 m R^{2}}} = \frac{2}{3} g \sin θ$

Now $l = \frac{1}{2} a_{2} t_{2}^{2} \Rightarrow t_{2} = \sqrt{\frac{2 l}{\frac{2}{3} g \sin θ}}$

   $\Rightarrow t_{2} = \sqrt{\frac{3}{2}} \sqrt{\frac{2 l}{g \sin θ}} = \sqrt{\frac{3}{2}} t$ $\dots (i i)$

On comparing, $\sqrt{\frac{3}{2}} t = \sqrt{\frac{α}{2}} t \Rightarrow α = 3$

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