A skater has rotational inertia 4.2 kg-m2 with his fists held to his chest and 5.7 kg?m2 with his arms outstretched. The skater
is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they�re essentially on his rotation axis, how fast will he be spinning? Express your answer using two significant figures. ?f=
1 answer:
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Answer:
Counterclockwise
Explanation:
We need to calculate the clocwkise and the counterclockwise torque.
The clockwise torque is the one generated by the child sitting on the right. This torque is given by:
where
is the force exerted by the child (his weight)
d = 1.8 m is the distance from the pivot point
So, the clockwise torque is
The counterclockwise torque is the one generated by the adult pushing on the left, and it is given by
where
F = 151 N is the force applied
d = 3.0 m is the distance from the pivot
Substituting,
So, the net torque is
And since the counterclockwise momentum is greater than the clockwise one, the teeter-totter will rotate counterclockwise.