A uniform rod of length 8 a and mass 6 m lies on a smooth horizontal surface. Two-point masses m and 2 m moving in the same plane with speed 2 v and v respectively strike the rod perpendicular at dist

Question

A uniform rod of length 8 a and mass 6 m lies on a smooth horizontal surface. Two-point masses m and 2 m moving in the same plane with speed 2 v and v respectively strike the rod perpendicular at distances a and 2 a from the midpoint of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is

TestHub · Accepted Answer

Applying conservation of angular momentum about point $O$ ,

$m (a) (2 v) + 2 m (2 a) (v) = I ω$

or $ω = \frac{6 m a v}{I}$ ...(i)

Now,

$I = \frac{6 m {(8 a)}^{2}}{12} + m {(a)}^{2} + 2 m {(2 a)}^{2}$

$= 32 m a^{2} + m a^{2} + 8 m a^{2} = 41 m a^{2}$

Hence, from Eq.(i)

$ω = \frac{6 m a v}{41 m a^{2}} = \frac{6 v}{41 a}$

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