A bead P slides on a frictionless semi-circul | vedclass

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Question

(B) Let the radius of the semi-circle be $R$. The horizontal distance covered by bead $Q$ to reach point $B$ is $2R$. Since it moves with a constant h

Vedclass · Accepted Answer

$t_{P} < t_{Q}$

Vedclass · Answer

(B) Let the radius of the semi-circle be $R$. The horizontal distance covered by bead $Q$ to reach point $B$ is $2R$. Since it moves with a constant horizontal velocity $v$, the time taken is $t_Q = \frac{2R}{v}$.For bead $P$, the horizontal component of velocity is given as $v_x = v$ at point $S$. As the bead slides down the frictionless semi-circular wire, its speed increases due to gravity. The horizontal component of its velocity $v_x$ will be $v \cos 	heta$, where $	heta$ is the angle the velocity vector makes with the horizontal. Since the bead is constrained to move along the semi-circle, its horizontal velocity component $v_x$ will always be greater than or equal to its initial horizontal velocity $v$ as it moves towards $C$ and then towards $B$. Because the average horizontal velocity of $P$ is greater than $v$, and the total horizontal distance $P$ needs to cover is $2R$, the time taken $t_P$ must be less than $t_Q$.

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