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Question

A smooth uniform rod of length L and mass M has two identical beads of negligible size, each of mass m, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity ω0 about an axis perpendicular to the rod and passing through the mid-point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is


A
ω=mω0m+6M
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B
ω=Mω06M+m
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C
ω=6mω0M+6m
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D
ω=Mω0M+6m
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Solution

The correct option is D ω=Mω0M+6m
Consider the rod and two beads together as a system, there are no external torques on the system about the fixed fixed point O (because there are no external forces ). Hence, angular momentum of the system about the fixed point O is conserved. Initially, two beads of mass m each are at the centre of the rod of mass M and length L, and angular velocity ω0.


Initial angular momentum of the system about axis of rotation is,
L0=I0ω0
where I0=ML212+m(0)2+m(0)2
i.e L0=ML212ω0

Finally, the two beads are at the extreme ends of the rod rotating with an angular velocity ω. Thus, the final angular momentum of the system about the axis of rotation is
L=Iω
where I=[ML212+m(L2)2+m(L2)2]
i.e L=[M12+m2]L2ω
From conservation of angular momentum, about the fixed point O
L0=L
ML212×ω0=[M12+m2]L2ω
ω=Mω0M+6m

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