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Question

A horizontal rod of length l=2 m and mass 2 kg with the balls A and B of mass 1 kg each at its ends is clamped at the centre C in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of mass 1 kg is dropped from a height of 5 m on the ball B. The particle collides with B and sticks to it. Find the angular velocity of the system just after the collision.


A
103 rad/s
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B
203 rad/s
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C
1011 rad/s
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D
3011 rad/s
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Solution

The correct option is D 3011 rad/s
Given,
Mass of ball A,B and P=1 kg
Mass of rod =M=2 kg
Length of rod =l=2 m

Using work energy theorem for particle P:
mgh+0=12mv2+0
v=2gh ... (1) is the speed of particle P just before collision :

There is no external torque on the system. Hence the angular momentum of the system remains same.
Hence, Li=Lf


Li=mvr
where, r=(l2)
Hence, Li=m(2gh)(l2)
(r is the perpendicular distance of velocity vector from the axis of rotation)


Lf=Iω
where I= MOI of the system
I=Ml212+3m(l2)2
I=Ml212+3ml24
Lf=(Ml212+3ml24)ω
From Li=Lf
m(2gh)×(l2)=(Ml212+3ml24)ω
1×2×10×5×(22)=(2×2212+3×1×224)×ω
10=(23+3)ω
i.e ω=3011 rad/s

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