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 force that the rod exerts on the stone 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

At the bottom of the circle, the force exerted by the rod on the stone is the sum of the gravitational force and the centripetal force required to keep the stone moving in a circle. The gravitational force is 20 N (mg = 2 kg * 10 m/s²). The centripetal force is calculated using mv²/r, where v is the speed at the bottom. Using energy conservation, potential energy at the top equals kinetic energy at the bottom: mgh = 0.5mv². Solving gives v² = 2gh = 24 m²/s². Thus, centripetal force is (2 kg * 24 m²/s²) / 1.2 m = 40 N. Total force = 20 N + 40 N = 60 N.

A Small but Dense 2