Rotational Motion Question 209
Question: A child with mass m is standing at the edge of a playground merry-go-round (A large uniform disc which rotates in horizontal plane about a fixed vertical axis in parks) with moment of inertia Vs I, radius R, and initial angular velocity w. The child jumps off the edge of the merry-go-round with a velocity v with respect to the ground in direction tangent to periphery of the disc as shown. The new angular veloity of the merry-go-round is :
Options:
A) $\sqrt{\frac{I{{\omega }^{2}}-mv^{2}}{I}}$
B) $\sqrt{\frac{(I+mR^{2}){{\omega }^{2}}-mv^{2}}{I}}$
C) $\frac{I\omega -mvR}{I}$
D) $\frac{(I\omega -mR^{2})\omega -mvR}{I}$
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Answer:
Correct Answer: D
Solution:
[d] Let the angular velocity of disc after child jumps off.
be $\omega ‘$
$\therefore $ From conservation of angular momentum $(I+mR^{2})\omega =mvR+I\omega $.
$\therefore \omega ‘=\frac{(I+mR^{2})\omega +mvR}{I}$
