# Homework Help: Help!: A ball attached to two strings (circular mtoion)

1. Dec 16, 2007

### Cate

1. The problem statement, all variables and given/known data
A ball of mass 1.21 kg is attached to a rotating shaft by two strings, putting the ball into circular motion. Tension on the upper string is 35N. (The length of the strings is 1.6m)

a) draw a free body diagram for forces acting on the ball.
b) determine the tension in the lower string
c) determine the net force acting on the ball
d) determine the speed of the ball

2. Relevant equations

ac= v^2/r

Fc=mac

3. The attempt at a solution

My problem is with a) I always have a hard time doing free body diagrams and breaking down the x and y components. Once I get that I know how to solve for c) and d). Some direction would be much appreciated!

2. Dec 16, 2007

### CaptainZappo

If the ball is moving in uniform circular motion, is there any net force in the vertical direction? Perhaps you can use this to your advantage.

3. Dec 16, 2007

### Cate

yes i assumed that the fnet in the y direction is zero. but what about the two strings?

4. Dec 16, 2007

### Cate

could someone please point me in the right direction as to find b)?

5. Dec 16, 2007

### Staff: Mentor

I assume the ball is traveling in a horizontal circle? A diagram would be helpful.

The strings apply their tension forces to the ball. Consider that the ball is centripetally accelerating and that the radial component of the net tension on the ball provides the centripetal force.

6. Dec 16, 2007

### Dick

The sum of the vertical components of the tensions must equal mg to keep the ball suspended. The sum of the horizontal components must equal mv^2/r. But I don't see any way to actually get a number for anything unless you know the angle between the strings or the radius of rotation r.

7. Dec 16, 2007

### Cate

I assume the radius of rotation would be 1.6m

8. Dec 16, 2007

### Staff: Mentor

Isn't that the length of the strings? (Total length? Or length of each?) Are the strings horizontal?

9. Dec 16, 2007

### Cate

yes, 1.6 m is the length og the strings (individual) but it's also the disatnce between the two strings on the rod

10. Dec 16, 2007

### Cate

The diagram looks like #14 at this web page

http://teachers.oregon.k12.wi.us/fishwild/APPhysics/ForceMotion2.pdf [Broken]

Last edited by a moderator: May 3, 2017
11. Dec 16, 2007

### Staff: Mentor

Apply a little triangle geometry and you can figure out the radius of the ball's circular path.

12. Dec 16, 2007

### Cate

what do you mean trianle geometry? why isn't the radius 1.6m? how will this help me solve for the lower tension

13. Dec 16, 2007

### Staff: Mentor

If the string is 1.6m and it is at some angle with the horizontal, how can the radius also be 1.6m?

Draw yourself a diagram. The rod and the two strings form an equilateral triangle.

14. Dec 16, 2007

### Cate

o.k so if it's an equaliteral trangle then all three sides are 1.6m so i'm trying to find the height of the triangle?

15. Dec 16, 2007

### Staff: Mentor

That's right. Or just make a right triangle by drawing in the radius. (What are the angles of an equilateral triangle?)

16. Dec 16, 2007

### Cate

17. Dec 16, 2007

### Staff: Mentor

Looks good.

18. Dec 16, 2007

### Cate

thanks for your help, but could you give me a hint has to how to solve for the lower tension?

19. Dec 16, 2007

### Staff: Mentor

What's the net horizontal force on the ball? (Call the lower tension T.) That force is the centripetal force.