A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long.The other end of the rod is attached to a frictionless pivot.The rod is raised until it is vertical,with the stone above the pivot.The rod is released and the stone moves in a vertical circle with no air resistance.What is the tension in the rod as the stone moves through the bottom of the circle?
A)20 N
B)40 N
C)60 N
D)80 N
E)100 N

Question

Quizplus · Accepted Answer

The tension at the bottom of the circle is the sum of the gravitational force and the centripetal force required to keep the stone moving in a circle. The gravitational force is 2.0 kg * 9.8 m/s² = 19.6 N. The centripetal force is calculated using the formula mv²/r. At the bottom, the potential energy at the top is converted to kinetic energy: mgh = 0.5mv², where h = 1.2 m. Solving for v² gives v² = 2gh = 2 * 9.8 * 1.2 = 23.52 m²/s². The centripetal force is then (2.0 kg * 23.52 m²/s²) / 1.2 m = 39.2 N. The total tension is 19.6 N + 39.2 N = 58.8 N, which rounds to 60 N.

A Small but Dense 2