JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 7)
A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity $v_0$ at the point ' A ', the string becomes slack when, the bob reaches at the point ' $D$ '. The ratio of the kinetic energy of the bob at the points B and C is _________.
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Explanation
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As the string becomes slack when the bob reaches at the point D so tension is zero at this point. And the bob just completes the circle.
We know, to complete a vertical loop of radius R, the minimum velocity required at the bottom of loop is $\sqrt{5gR}$.
So, $${V_0} = \sqrt {5gl} = {V_A}$$ (as R = $l$)
So, using energy conservation,
$${1 \over 2}mv_A^2 = {1 \over 2}mv_B^2 + mgh$$
$$ \Rightarrow {1 \over 2}m(5gl) = {1 \over 2}mv_B^2 + mg\left( {{l \over 2}} \right)$$
$$ \Rightarrow K{E_B} = {{mgl} \over 2}(5 - 1) = 2mgl$$
For point C,
$${1 \over 2}mv_A^2 = {1 \over 2}mv_c^2 + mg\left( {l + {l \over 2}} \right)$$
$$ \Rightarrow {5 \over 2}mgl = K{E_C} + {{3mgl} \over 2}$$
$$K{E_C} = {{mgl} \over 2}(5 - 3) = mgl$$
So, $${{K{E_B}} \over {K{E_C}}} = {{2mgl} \over {mgl}} = 2$$
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