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Rotational Work and Energy

A particle of...

Question

A particle of mass $m$ attached to a string of length $l$ is describing circular motion on a smooth plane inclined at an angle $α$ with the horizontal. For the particle to complete the loop, its velocity at the lowest point should exceed

A

$\sqrt{5 g l}$

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B

$\sqrt{5 g l (c o s α + 1)}$

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C

$\sqrt{5 g l t a n α}$

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D

$\sqrt{5 g l s i n α}$

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Solution

The correct option is D
$\sqrt{5 g l s i n α}$

A is the lowest point and B the highest point.
At B, in critical case tension is zero.
Let velocity of particle at B at this instant be $v_{B}$ .
Then from Free Body diagram

$m g s i n α = \frac{m v_{B}^{2}}{l}$
or $v_{B}^{2} = g l s i n α$
Now as $h = 2 l s i n α$
$v_{A}^{2} = v_{B}^{2} + 2 g h = (g l s i n α) + 2 g (2 l s i n α$

therefore $v_{A} = \sqrt{5 g l s i n α}$

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