Question

# A uniform rectangular plate of mass m which is free to rotate about the smooth vertical hinge passing through the center and perpendicular to the plate is lying on a smooth horizontal surface. A particle of mass m moving with speed 'u' collides with the plate and sticks to it as shown in the figure. The angular velocity of the plate after the collision will be

A
12u5a
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B
12u19a
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C
3u2a
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D
3u5a
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Solution

## The correct option is D 3u5aAngular momentum of the system above hinge should be.Angular momentum of particle of mass m before collision is,Lp=a.muLp=muaLet after collision the angular velocity of the attached system be ωMoment of inertia about a perpendicular axis passing through hinge isI= moment of inertia of particle of mass m about hinge + Moment of inertia of the plate about hinge.m.(√5a2)+112m(a2+(2a)2)=54ma2+512ma2=2012ma2Applying conservation of angular mamentum we getIw=Lp=53ma253ma2.ω=mua⇒3u5a

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