can someone help me please?

Question

II don't get this problem. I don't know what the difference is between the ball at the top of its path and when it is at the bottom of its path.

The answer that I got was .000137. I think that is wrong. can someone show and explain to me how I do this?

A ball on the end of a string is cleverly revolved at a uniform rate in a vertical circle of radius 85.0 cm. Its speed is 3.90 m/s and its mass is 0.300 kg.

What is the tension in the string when the ball is at the top of its path? What about when it is at the bottom of its path.

Answer 1

What may be confusing you is the clever way the ball maintains a constant speed of 3.9 m/s

What I suggest is to consider the two problems independently.

At the top of its path is when the ball is at the highest vertical height it will reach. In essence, the question is structured so that you consider the upward force due to angular speed pulling agianst the string. Gravity will pull downward, which will make the tension less.

Let's first compute the force due to angular speed:

F=m*v^2/r and points outward from the center of the circle

for this specific case:
.3*3.9^2/.85
=5.37 N

At the top, gravity will offset this so tension is
5.37-.3*9.81
=2.43 N

at the bottom gravity will add additional tension in the string since the angular force and gravity now point in the same direction
=5.37+.3*9.81
=8.31 N

j