A 0.500 kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long.The stone is moving around the path at a constant frequency of 2.20 rev/sec.At the moment the stone is overhead,the stone is released.The magnitude and direction of the velocity of the stone when it leaves the circular path is
A) 10.4 m/s horizontal.
B) 10.4 m/s vertical.
C) 22.0 m/s horizontal.
D) 22.0 m/s vertical.
E) 31.4 m/s horizontal.

Question

Quizplus · Accepted Answer

The velocity of the stone when released is the tangential velocity, which is horizontal at the top of the circle. The tangential velocity is calculated using v = 2πr * frequency, where r = 0.75 m and frequency = 2.20 rev/sec. This gives v = 10.4 m/s.

A 0500 Kg Stone Is Moving in a Vertical Circular Path