- #1

littlebearrrr

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## Homework Statement

A ball of mass 125g is attached to a string .900 meters long. It is then set into vertical circular motion with 38 RPM. What is the tension of the string at the top of the circle and at the bottom of the circle?

## Homework Equations

∑Fy = may = marad

arad = v

^{2}/R = 4∏

^{2}R/T

^{2}

## The Attempt at a Solution

*First, I found T (period) by using the given RPM:*

(38 rev/1 min)(1 min/60 s) = 0.633 rev/s -->

**1.58 s per revolution = T**

*Used T to calculate arad (radial acceleration):*

arad = 4∏

^{2}(.900 m) / (1.58 s)

^{2}=

**14.23 m/s**

^{2}*Finding the tension at the top of the circle:*

The only forces acting on the ball at the top are its weight and the tension force acting in the same direction (downward). With these facts, I apply Newton's second law for the vertical direction and solve for the tension -

∑Fy = may = marad = w + T

T = marad - w = (.125 kg)(14.23 m/s

^{2}) - (.125 kg)(9.81 m/s

^{2}) =

**0.553 N**

*Finding the tension at the bottom of the circle:*

At the bottom, the forces now oppose each other. The tension is directed upward, while the weight remains directed down. Again, I apply Newton's second law -

∑Fy = may = marad = T - w

T = marad + w = (.125 kg)(14.23 m/s

^{2}) + (.125 kg)(9.81 m/s

^{2}) =

**3.01 N**

Just checking to see if this is correct. Thank you in advance!

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