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Standard XII

Physics

Work Energy Theorem Application

A ring of rad...

Question

A ring of radius $R$ lies in vertical plane. A bead of mass $m$ can move along the ring without friction. Initially the bead is at rest at the bottom most point on ring. The minimum constant horizontal speed $v$ , with which the ring must be pulled such that the bead completes the vertical circle, is:

\sqrt{3 g R}

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\sqrt{4 g R}

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\sqrt{5 g R}

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\sqrt{5.5 g R}

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Solution

The correct option is B $\sqrt{4 g R}$
In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is $v$ at the lowest position. The condition for bead to complete the vertical circle is, its speed at top position, $v_{t o p} \geq 0$
So, to just complete the full circle, $v_{t o p} = 0$
From conservation of energy,
$\frac{1}{2} m v_{t o p}^{2} + m g (2 R) = \frac{1}{2} m v^{2}$
$v = \sqrt{4 g R}$

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A ring of radius R lies in vertical plane. A bead of mass m can move along the ring without friction. Initially the bead is at rest at the bottom most point on ring. The minimum constant horizontal speed v, with which the ring must be pulled such that the bead completes the vertical circle, is: