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Standard XII

Physics

Law of Conservation of Mechanical Energy

A particle of...

Question

A particle of mass $m$ is fixed to one end of a light rigid rod of length $l$ and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

z e r o

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\sqrt{g l}

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\sqrt{1.5 g l}

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\sqrt{2 g l}

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Solution

The correct option is A $z e r o$
Let the velocity at B be $v$ and at A be $\sqrt{4 g R}$ .
Energy conservation at A and B, $\frac{1}{2} m (\sqrt{4 g R})^{2} + 0 = \frac{1}{2} m v^{2} + m g (2 R)$
$⟹ v = 0$
Thus minimum velocity at highest point is Zero.

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A particle of mass m is fixed to one end of a light rigid rod of length l and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

The correct option is A zeroLet the velocity at B be v and at A be √4gR.Energy conservation at A and B, 12m(√4gR)2+0=12mv2+mg(2R)⟹v=0Thus minimum velocity at highest point is Zero.

A particle of mass $m$ is fixed to one end of a light rigid rod of length $l$ and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

The correct option is A $z e r o$
Let the velocity at B be $v$ and at A be $\sqrt{4 g R}$ .
Energy conservation at A and B, $\frac{1}{2} m (\sqrt{4 g R})^{2} + 0 = \frac{1}{2} m v^{2} + m g (2 R)$
$⟹ v = 0$
Thus minimum velocity at highest point is Zero.