A small block with mass 0.0400 kg slides in a vertical circle of radius 0.0700 m 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 3.85 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

Question

A small block with mass 0.0400 kg slides in a vertical circle of radius 0.0700 m 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 3.85 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

Answer 1

To find the magnitude of the normal force that the track exerts on the block when it is at the top of its path, we need to consider the forces acting on the block at that position.

At the top of the block's path, it experiences two forces: the gravitational force (mg) pulling it downward, and the normal force (N) exerted by the track on the block in the upward direction.

Since the block is moving in a vertical circle, it experiences uniform circular motion, which means its acceleration (centripetal acceleration) points towards the center of the circle.

Now, let's break down the forces acting on the block at the top of its path:

1. Gravitational force (mg): The gravitational force always acts vertically downward and has a magnitude of mg, where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).

2. Normal force (N): The normal force acts perpendicular to the track's surface. At the top of the path, the normal force is directed towards the center of the circle because it is required to supply the inward centripetal force needed to keep the block moving in a circle.

Since the block is not in contact with any other surface (except the track), there are no other forces present that affect the normal force.

Now, the centripetal force required for the block to move in a circle is given by the equation:

Fc = (mass × velocity^2) / radius

At the top of the path, the velocity of the block will be minimum and equal to zero. Therefore, the centripetal force required to keep the block moving in a circle is also zero.

So, the normal force (N) at the top of the path must be equal to the gravitational force (mg):

N = mg

Substituting the given values:

mass (m) = 0.0400 kg
gravitational acceleration (g) = 9.8 m/s^2

N = (0.0400 kg) × (9.8 m/s^2)
N = 0.392 N

The magnitude of the normal force that the track exerts on the block when it is at the top of its path is 0.392 N.