Question

A small block with mass 0.0475kg slides in a vertical circle of radius 0.0770m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.90N

Part A

What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

Express your answer with the appropriate units.

Answer 1

The easiest way to approach the problem is understanding that when the block is at the bottom of the circle, there are two forces acting on it: gravity and the normal force. The summation of these two forces, if done properly, should yield the net force on the block and should equal to the centripetal force as it keeps the block in circular rotation. As such:

F (centripetal) = F (Normal) - F (Gravity)

               

                          = (3.9 N) - (.0475 kg) ( 9.81 m/s^2 )

                          = 3.434025 N

Now, we know the centripetal force ( or the net force) necessary for the block to stay in circular motion. Using this, we can calculate the required force the track exerts at the top, or the normal force at the top. Note that this time, all of the forces are downward, so all are negative.

- F (centripetal) = - F (Normal) - F (Gravity)

- 3.434025 N = - F (normal) - (.0475 kg) ( 9.81 m/s^2 )

F (normal) = 2.96805 N

F (normal) = 2.97 N downward