# Homework Help: Ball Pendulum Problem

1. Nov 2, 2008

### blayman5

1. The problem statement, all variables and given/known data
A heavy ball swings on a string in a circular
The two highest points of the ball’s trajec-
tory are Q and Q′; at these points the string is
±28◦ from the vertical. Point P is the lowest
point of the ball’s trajectory where the string
hangs vertically down.
The acceleration of gravity is 9.8 m/s2.
What is the magnitude of the ball’s acceler-
ation at the point Q? Answer in units of
m/s2.

3. The attempt at a solution
Wouldn't the acceleration at Q be 0 because there is no velocity at that point.

2. Nov 2, 2008

### Staff: Mentor

The centripetal acceleration will be zero, but what about the tangential acceleration?

3. Nov 2, 2008

### blayman5

Ah, there would be tangential acceleration. How would you go about finding it? The initial steps.

4. Nov 2, 2008

### Staff: Mentor

Apply Newton's 2nd law. What forces act on the ball? What are their tangential components?

5. Nov 2, 2008

### blayman5

The forces acting on the ball are the tension and the mg.
TcosO=mg+ma. Since I do not know the tension, how can I solve for tangential acceleration?

6. Nov 2, 2008

### Staff: Mentor

Does the tension have a tangential component?

7. Nov 2, 2008

T=mg

mgcos0=ma
a=gcos0?

8. Nov 2, 2008

### blayman5

I did TcosO-mg=ma
T=mg+ma/cos0

but there is no mass given in the problem

9. Nov 2, 2008

### Staff: Mentor

Do this. Draw a careful diagram of the ball and string showing all forces acting on the ball. (There are only two.) Then find the tangential component of each force. (Tangential, in this case, is perpendicular to the string.)

You won't need the mass to find the tangential acceleration. Just call it "m"--it will drop out when you apply Newton's 2nd law.

10. Nov 2, 2008

### blayman5

oh ok i did the process and i got a=gtan0.
If the string broke at point Q, would it undergo projectile motion with a horizontal initial velocity?

11. Nov 3, 2008

### Staff: Mentor

That is not correct.
What's the speed of the ball at point Q?

12. Nov 3, 2008

### blayman5

there is the tension and its components and mg as the forces on the ball. The component TsinO would be the tangential force. Tcos0=mg. T=mg/cos0 Tsin0=ma (mg/cos0)(sin0)=ma
Where did I misinterpret the problem?

The velocity at point Q is 0m/s

13. Nov 3, 2008

### Staff: Mentor

Right. String tension and mg are the only forces on the ball.
No, Tsinθ is the horizontal component of the tension.
No, Tcosθ (the vertical component of tension) does not equal mg. It would if the ball were in equilibrium, but it's not.
Find the tangential components of the force on the ball.
Right. So if the ball were let go at that point (the string was cut), what would its motion look like?

14. Nov 4, 2008

### blayman5

ok i got gsin0 and it was correct.
Thanks