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 speed 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.

Question

Quizplus · Accepted Answer

The speed of the stone when it leaves the circular path can be found using the formula for the velocity of an object moving in a circle with a constant frequency, $v = 2\pi r f$, where $r$ is the radius of the circle and $f$ is the frequency of the rotation. Given $r = 0.75$ m (since the length of the string is the radius of the circular path) and $f = 1.50$ rev/sec, we can substitute these values into the formula:$$v = 2\pi (0.75) (1.50) = 7.07 \, \text{m/s}$$Therefore, the speed of the stone when it leaves the circular path is 7.07 m/s.

A 0500 Kg Stone Is Moving in a Vertical Circular Path