A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B, perpendicular to the rod in the horizontal direction. Speed of particle P at a distance l/6 from the centre towards A of the rod after time t = πml 12J is
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Question

A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B, perpendicular to the rod in the horizontal direction. Speed of particle P at a distance l/6 from the centre towards A of the rod after time t = πml 12J is

Options

G1 Physics · Accepted Answer

Impulse J at end B gives centre-of-mass speed VCM=J/m. Angular impulse about the centre is J(L/2)=Iω with I=mL²/12, so ω=6J/(mL). During t=πmL/(12J), the rod turns through ωt=π/2. Point P is at L/6 from the centre, so its rotational speed is ωL/6=J/m, perpendicular to VCM; hence vP=√[(J/m)²+(J/m)²]=√2J/m. Therefore option D.