# Circular motion: Finding tension and theta

swede5670

## Homework Statement

Estimate the force a person must exert on a string attached to a 0.200 kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 1.40 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the angle θ it makes with the horizontal. [Hint: Set the horizontal component of FT equal to maR; also, since there is no vertical motion, what can you say about the vertical component of FT?]

Fnet=ma
w=v/2(pi)(r)
so
v=w*2(pi)(r)

## The Attempt at a Solution

I've made a free body diagram but I was not shown how to solve any problems like this so I'm a little unsure as to how I'm going to begin. I decided to break up tension into horizontal and vertical components
horizontal tension = cos (theta) * t
vertical tension = sin (theta) * t
Then I found the acceleration by converting the revolutions per second into m/s and then finally finding the acceleration which is 1.4*(2*pi*.6)=5.277 then i converted that into acceleration since fnet=ma. v^2/r=a so 27.84/.600=46.41m/s/s
I then take this acceleration and multiply it by mass which is .2kg and I get fnet=9.28.
And here's where I become confused. I'm trying to solve for tension so I can't get rid of that variable but I'm fairly sure I need to substitute something in here for the equation to work.
T*sine(theta)=9.28

Since the ball is not moving up or down we can say that
t*cosine(theta)-mg=0
which ends up as
t*cos(theta)=1.962

I have 2 variables (tension and theta) and I'm not sure where/how to substitute for this problem.

Homework Helper
m*v^2/r = FT*cos theta...(1)
mg=FT*sin theta...(2)
Divide (2) by (1) and find the angle.

swede5670
I'm not sure how to do that, I understand both of the equations but how do you divide one equation by the other? Are you telling me to substitute?
And are you sure about your two equations? I have something similar in my first post but the sin and cos are switched
you said: m*v^2/r = FT*cos theta but I thought it would be m*v^2/r = FT*sin
The reason I thought it was sin here was because I'm using sine to find the horizontal component of tension, cos is the vertical component isn't it?
And on the second equation mg=FT*sin theta I said it was mg=ft*cos (theta) because mg must equal the vertical component not the horizontal one.

swede5670
^^^^^ Bump

Homework Helper
I'm not sure how to do that, I understand both of the equations but how do you divide one equation by the other?

Since they are equations it's like dividing each side by the same thing isn't it?

This will yield the angle θ as the tangent or cotangent of θ depending on which way you do it.

Armed with the angle can't you determine the FT?

swede5670
Since they are equations it's like dividing each side by the same thing isn't it?

This will yield the angle θ as the tangent or cotangent of θ depending on which way you do it.

Armed with the angle can't you determine the FT?

9.28= FT*sin (theta)

1.96200=ft*cos (theta)

so Ft cancels and i divide 9.28/1.962=4.72986748
and then tan (4.72986748) to get theta?

The problem is that only gives me .0827 and i have a feeling that is not what my angle should be.

swede5670
Wait, I have to use tan^-1 right? if that's the case I get 78.06 degrees which seems a lot more reasonable. However, I checked and the answer is wrong, does anyone know what I did incorrectly?

Homework Helper
That is right.

swede5670
Well the website I had to submit it to said it was wrong, so maybe it's wrong?

Mentor
9.28= FT*sin (theta)

1.96200=ft*cos (theta)
Where did you get those numbers?

swede5670
I was trying to find acceleration when I found 9.28
1.4*(2*pi*.6)=5.277 m/s (velocity)
v^2/r=a so 27.84/.600=46.41m/s/s (acceleration)
and I found it when I put it in the Fnet=ma equation
I mutliplied mass by the acceleration I had found (46.41*.20) which equals 9.28

I found 1.962 as mg, the mass was .2 kg and multiplied it by the acceleration of gravity which is 1.962

Mentor
You should be using the equations provided by rl.bhat in post #2.