JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 1)
A bead of mass ' $m$ ' slides without friction on the wall of a vertical circular hoop of radius ' $R$ ' as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is ' $R$ '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' $R$ ', would be (spring constant is ' $k$ ', $g$ is accleration due to gravity)
_28th_January_Morning_Shift_en_1_1.png)
Explanation
_28th_January_Morning_Shift_en_1_2.png)
When mass is at position A,
Elongation of spring, $${x_1} = 2R - R = R$$
(As equilibrium length of spring is R)
$${h_1} = 2R$$
$${v_1} = 0$$ (As the bead is released from zero initial speed)
When mass is at position B,
$${x_2} = R - R = 0$$
$${h_2} = R/2,{v_2} = v$$
So using energy conservation,
$${E_1}^\circ = {E_f}$$
$$ \Rightarrow {1 \over 2}kx_1^2 + {1 \over 2}mv_1^2 + mg{h_1} = {1 \over 2}kx_2^2 + {1 \over 2}mv_2^2 + mg{h_2}$$
$$ \Rightarrow {1 \over 2}k{R^2} + mg(2R) = {1 \over 2}k(o) + {1 \over 2}m{v^2} + mg{R \over 2}$$
$$ \Rightarrow {1 \over 2}k{R^2} + 2mgR - {{mgR} \over 2} = {1 \over 2}m{v^2}$$
$$ \Rightarrow {1 \over 2}m{v^2} = {1 \over 2}k{R^2} + {{3mgR} \over 2}$$
$$ \Rightarrow {v^2} = {{k{R^2}} \over m} + 3gR$$
$$ \Rightarrow v = \sqrt {{{k{R^2}} \over m} + 3gR} $$
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