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Standard XII

Physics

Throwing and Catching at Same Height

A small ring ...

Question

A small ring of mass $m$ attached at an end of a light string of length $l = 0.6 m$ the other end of which is tied to a block $B$ of mass $2 m$ . The ring is free to move on a fixed smooth horizontal rod. Find the velocity of A ring and tension in the string when it becomes vertical.

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Solution

From energy Conservation
$\frac{1}{θ} m v^{2} = m g l + \frac{m g l}{3}$ (where $\frac{m}{3}$ is the effective mass of string)
$\frac{1}{θ} m v^{2} = \frac{4}{3} m g l$
$v^{2} = \frac{8 g l}{3}$
$v = \sqrt{\frac{8 g l}{3}}$
$∴ x = 8$

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A small ring of mass m attached at an end of a light string of length l=0.6m the other end of which is tied to a block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of A ring and tension in the string when it becomes vertical.

From energy Conservation1θmv2=mgl+mgl3 (where m3 is the effective mass of string)1θmv2=43mglv2=8 gl3v=√8 gl3∴x=8

A small ring of mass $m$ attached at an end of a light string of length $l = 0.6 m$ the other end of which is tied to a block $B$ of mass $2 m$ . The ring is free to move on a fixed smooth horizontal rod. Find the velocity of A ring and tension in the string when it becomes vertical.

From energy Conservation
$\frac{1}{θ} m v^{2} = m g l + \frac{m g l}{3}$ (where $\frac{m}{3}$ is the effective mass of string)
$\frac{1}{θ} m v^{2} = \frac{4}{3} m g l$
$v^{2} = \frac{8 g l}{3}$
$v = \sqrt{\frac{8 g l}{3}}$
$∴ x = 8$