Physics Rotation2?

Moment of inertia of the bug about the pivot (i.e. the other end of the bar)
= 0.25 x 1.8^2 g.cm^2 = 0.81 g.cm^2
Moment of inertia of the bar about the pivot
= (1 x 1.8^2)/3 g.cm^2 = 1.08 g.cm^2

Angular velocity of the bug when it jumps off the bar
= 3.9/1.8 rad/s = 2.167 rad/s

Using Law of Conservation of Angular Momentum,
Initial angular moment of system (bug + bar) = 0
Final angular momentum of system = (0.81 x 2.167) + 1.08w
where w is the angular velocity of the bar just after the bug jumps off.

Hence, 0 = (0.81 x 2.167) + 1.08w
i.e. w = -1.625 rad/s (the -ve sign indicates the rotational direction of the bar, which is opposite to the jumping direction of the bug)

Therefore, the angular speed of the bar is 1.625 rad/s

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