Question

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 1.50 rev/sec. At the moment the stone is overhead, the stone is released. The velocity of the stone when it leaves the circular path is
A. 5.55 m/s.
B. 7.07 m/s.
C. 7.75 m/s.
D. 8.35 m/s.
E. 9.00 m/s.

Answer 1

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, $v$, which is given by

$v=\omega r$

where $\omega$ is the angular speed and $r$ is the radius of the circular path.

$\omega$ is given by

$\omega = 2\pi f$

where $f$ is the frequency of revolution.

Thus

$v=2\pi fr$

Using values from the question,

$v=2\pi\times 1.50\times0.75$

Note the conversion of 75 cm to 0.75 m

$v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07$