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Question

A circular wooden loop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of mass m, and moving with a velocity V, strikes the loop at the bottommost point and gets embedded in it. The thickness of the loop is much smaller than R. The angular velocity with which the system rotates just after the bullet strikes the loop is


A
V3R
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B
3V4R
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C
V4R
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D
2V3R
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Solution

The correct option is A V3R
Assume that just after the collision, velocity of COM of the system is VCOM and angular velocity about COM is ω.


Just after collision:
Distance of COM of the system from O.
yCOM=m1y1+m2y2m1+m2
yCOM=m(0)+m(R)m+m=R2
There is no external force on the system. Hence linear momentum will be conserved.
mV+0=(m+m)VCOM
VCOM=V2 ... (1)
There is no torque on the system. Hence angular momentum will be conserved. Thus about COM,
Li=Lf
where Li=mv(R2)
Lf=LTranslational+LRotational
Here, LTranslational=0 {because VCOM passing through COM}
i.e Lf=ICOMω
MOI of system (loop + bullet) about COM
ICOM=(Iloop)COM+(Ibullet)COM
=(mR2+m(R2)2)+m(R2)2
ICOM=3mR22
Therefore, Li=Lf
mV(R2)=(32mR2).ω
ω=V3R

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