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.

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. Initially, the stone has kinetic energy and potential energy due to its height. At the moment it is released, all of the potential energy is converted to kinetic energy.

The potential energy at the top of the circle is given by:
PE = mgh
where m is the mass of the stone, g is the acceleration due to gravity, and h is the height of the stone above some reference point (usually taken to be the lowest point in the circle). In this case, h = 0 and the potential energy is zero.

The kinetic energy of the stone is given by:
KE = (1/2)mv^2
where v is the speed of the stone.

At the bottom of the circle, the speed of the stone is given by:
v = sqrt(2gh)
where h is now the height of the stone above the lowest point in the circle. In this case, h = 2R (where R is the radius of the circle, which is equal to 0.75 m), so we have:
v = sqrt(2g(2R))
v = sqrt(2(9.81 m/s^2)(1.5 m))
v = 7.07 m/s

Therefore, the velocity of the stone when it leaves the circular path is 7.07 m/s, which is answer choice B.

A 0500 Kg Stone Is Moving in a Vertical Circular Path