A meter scale is suspended freely from one of its ends. Its another end is given a horrizontal velocity v such that it completes one revolution in the vertical circle . The value of v is?

Question

Eshan · Answer

Dear student,

For the meter scale to complete a revolution, the scale just needs to complete half the circle to rise vertically until the scale is inverted since after that the scale can fall under the effect of gravity only.

From initial state to the final state, the total energy should be conserved.
The angular velocity of the rod will be $\omega=\dfrac{v}{l}$

Hence $\dfrac{1}{2}(m\dfrac{l^2}{3})\omega^2=mgl$

$\implies \omega=\sqrt{\dfrac{6g}{l}}$
$\implies v=\sqrt{6gl}$

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A meter scale is suspended freely from one of its ends. Its another end is given a horrizontal velocity v such that it completes one revolution in the vertical circle . The value of v is?

A meter scale is suspended freely from one of its ends. Its another end is given a horrizontal velocity v such that it completes one revolution in the vertical circle . The value of v is?

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A meter scale is suspended freely from one of its ends. Its another end is given a horrizontal velocity v such that it completes one revolution in the vertical circle . The value of v is?

A meter scale is suspended freely from one of its ends. Its another end is given a horrizontal velocity v such that it completes one revolution in the vertical circle . The value of v is?

1 Answers

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