A block of mass m is released from rest at a height R above a horizontal surface. The acceleration due to gravity is g. The block slides along the inside of a frictionless circular hoop of radius R.
-What is the magnitude of the normal force exerted on the block by the hoop when the block reaches the bottom of the hoop?

Question

A block of mass m is released from rest at a height R above a horizontal surface. The acceleration due to gravity is g. The block slides along the inside of a frictionless circular hoop of radius R.

-What is the magnitude of the normal force exerted on the block by the hoop when the block reaches the bottom of the hoop?

Quizplus · Accepted Answer

At the bottom of the hoop, the normal force must provide the centripetal force needed to keep the block moving in a circular path. The normal force is the sum of the gravitational force (mg) and the centripetal force (mv^2/R), which is 2mg at the bottom. Therefore, the total normal force is 3mg.

A Block of Mass M Is Released from Rest at a Height