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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
296140_05e07b10600a4d3ba53c034ce4637a41.png

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 3u5a
Angular momentum of the system above hinge should be.
Angular momentum of particle of mass m before collision is,
Lp=a.mu
Lp=mua
Let after collision the angular velocity of the attached system be ω
Moment of inertia about a perpendicular axis passing through hinge is
I= 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
=2012ma2
Applying conservation of angular mamentum we get
Iw=Lp
=53ma2
53ma2.ω=mua
3u5a


978445_296140_ans_9a251a43e27a489eacec536d501a3986.png

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