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 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 it is released is tangential to the circular path and since it is released at the moment it is overhead, the velocity is horizontal. To find the velocity, we use the formula for the velocity of an object moving in a circle: $v = 2\pi r f$, where $r$ is the radius and $f$ is the frequency. Given $r = 0.75$ m and $f = 2.20$ rev/sec, converting revolutions per second to radians per second (since $1$ rev = $2\pi$ radians), we get $v = 2\pi \times 0.75 \times 2.20 = 10.4$ m/s.

A 0500 Kg Stone Is Moving in a Vertical Circular Path