# Homework Help: Circular Motion Problem (Using Newtonian Physics)

1. Sep 26, 2010

### Dougggggg

1. The problem statement, all variables and given/known data
The problem is basically a fair ride that has a pole as its base, another pole sticking out of it, the a cable with a chair at the end. I am given the length of the rod sticking out (3.00 m), the cable (5.00 m) and the angle between 30.0$$\circ$$. I have to find the time it takes for 1 revolution around. The only other figure I can come up with is just -g.

2. Relevant equations
$$\Sigma$$Fy=may=Fc-mgcos$$\theta$$=0
arad=$$\frac{v^2}{R}$$=$$\frac{4\pi^2}{T^2}$$

3. The attempt at a solution
All I have done so far is solved for the how far the seat is from that 30 degree angle in both x and y directions. I had been flying through these problems except this one is like a wall, I have been lost on it for the past about 4 hours. Though my frustration and anger has made it difficult to make any progress. If anyone can point me in the right direction, I see nothing but dead ends no matter which way I try to solve this. I am gonna try to rework my free body diagram, will edit when have more figured out.

Edit:Close thread, found out how to do this. Used some of forces equations to solve for m on both equations and then replaced my a with the second equation up there, my forces canceled, now just have some algebra to do.

Last edited: Sep 26, 2010
2. Sep 26, 2010

### MHrtz

Solve for centripetal acceleration.

3. Sep 26, 2010

### Dougggggg

I found out how to do it, I don't know if I could have solved for my centripetal acceleration since I have not been given any acceleration values other than gravitation. Maybe I am misunderstanding what you are telling me.

4. Sep 26, 2010

### MHrtz

How did you find the answer and do you know if it's right?

5. Sep 27, 2010

### Dougggggg

I simply broke down my free body diagram of the seat and cam up with these equations.

$$\Sigma$$Fsy = may = Fccos$$\theta$$-mg = 0
$$\Sigma$$Fsx = max = Fcsin$$\theta$$

Then moved the mg over to that 0, divided by g to solve for m. Then solved for m on the bottom equation and substituted for the m in each equation to have.

(Fcsin$$\theta$$)/ax = (Fccos$$\theta$$)/g

Then I used the second formula in my original post to replace my ax with the second part of the arad equation. Then had a big mess, however, my Fc ended up canceling and my only unknown variable left is T which is what I need to solve for XD.

I think I had more trouble with this simply because I was stressing over things going on in my life, and thus made it difficult to focus. I still appreciate all the help that was offered.

6. Sep 27, 2010

### MHrtz

The centripetal acceleration is ax^2 + ay^2 but you only needed the radial component ax. That's what I was getting at. Nonetheless, good work.

7. Sep 27, 2010

### Dougggggg

If anyone has completely worked this did you get an answer along the lines of 5.91 s? It is an even problem and my book only has odd answers in the back.