Question

A uniform rod of mass m and length | hinged at point O and a point mass m is fixed at its other end as shown in figure if rod is released from horizontal position, then what is the velocity of the point mass when rod becomes vertical?

ANSWER WITH EXPLANATION

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Answer 1

Solution :

⏭ Given:

✏ A uniform rod of mass m and length L hinged at point O and a point mass m is fixed at its other end.

✏ Rod is released from horizontal position.

⏭ To Find:

✏ Velocity of point mass when rod becomes vertical.

⏭ Concept:

✏ This question is completely based on concept of energy conservation.

⏭ Diagram:

✏ Please see the attached image for better understanding.

⏭ Calculation:

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$\implies \sf \: Distance \: between \: axis \: of \: rotation \\ \sf\:and \: COM \: of \: system = \red{ \frac{3L}{4}}$

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$\implies \sf \:As \: per \: parallel \: axis \: theorem \\ \\ \circ \sf\: M.I. \: related \: to \: point \: mass = \dfrac{m {L}^{2} }{3} + m {L}^{2} \\ \\ \circ \: \boxed{ \tt{ \green{I= \frac{4m {L}^{2} }{3}}}} \: \sf( wrt \: axis \: of \: rotation)$

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$\implies \sf \: \blue{U_i + K_i = U_f + K_f} \\ \\ \implies \sf \: \dfrac{2mg \times 3L}{4} + 0 = 0 + \frac{1}{2} I { \omega}^{2} \\ \\ \implies \sf \: \dfrac{3mgL}{2 } = \frac{1}{2} \times \dfrac{4m {L}^{2} }{3} \times \dfrac{ {v}^{2} }{ {L}^{2} } \\ \\ \implies \sf \: 3gL = \frac{4 {v}^{2} }{3} \\ \\ \implies \sf \: {v}^{2} = \frac{9}{4} \times gL\\ \\ \implies \: \underline{ \boxed{ \tt{ \large{ \orange{v = \frac{3}{2} \sqrt{gL}}}}}} \: \red{ \bigstar}$

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