First slide

Rotational motion

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 center of the rod and the system is rotating with angular velocity $ω_{0}$ 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

Moderate

A

$ω_{0}$
B

$\frac{M ω_{0}}{M + 12 m}$
C

$\frac{M ω_{0}}{M + 2 m}$
D

$\frac{M ω_{0}}{M + 6 m}$

Solution

Since there are no external forces, therefore, the angular momentum of the system remains constant.
Initially when the beads are at the centre of the rod angular momentum, $L_{1} = (\frac{M L^{2}}{12}) ω_{0} . . . . . . . . . . . (i)$
When the beads reach the ends of the rod, then angular momentum, $L_{2} = (m {(\frac{L}{2})}^{2} + m {(\frac{L}{2})}^{2} + \frac{M L^{2}}{12}) ω^{'} . . . . . . . . . . . . (i i)$
Equating (i) and (ii),
$\frac{M L^{2}}{12} ω_{0} = (\frac{m L^{2}}{2} + \frac{M L^{2}}{12}) ω^{'}$
$\Rightarrow ω^{'} = \frac{M ω_{0}}{M + 6 m}$

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