# Homework Help: Centripetal Force Ball Swing Question

1. Feb 14, 2013

### tcc88

1. The problem statement, all variables and given/known data
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.16 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.32 rev/s.

(a) Which rate of rotation gives the greater speed for the ball? [Which I got is 6.32]
(b) What is the centripetal acceleration of the ball at 8.16 rev/s? [?]
(c) What is the centripetal acceleration at 6.32 rev/s? [?]

2. Relevant equations
a[c] = v^2/r
T = 2∏r/v

3. The attempt at a solution

I know the I need to find the velocity to solve this question, but that is where I am having trouble. I am assuming 8.16 and 6.32 are the periods. So I get:
v = 2∏(0.300)/(8.16) for B; and then use this to find acceleration which is: a[c] = (0.23)^2/(0.300) = 1.176. But this is not the correct answer! Why?

2. Feb 14, 2013

### cepheid

Staff Emeritus
I think your answers for both part a and part b are wrong, and here's why:

The "rev/s: in 6.32 rev/s and 8.16 rev/s stands for "revolutions PER second." So it is most emphatically NOT a period. It tells you how many revolutions occur in one second. So, you might say, it tells you how often a revolution occurs. In fact, if you were inclined to phrase it differently, you might say that this number tells you how frequently a revolution occurs.

nudge nudge

3. Feb 14, 2013

### tcc88

So would the frequency be 1/8.16 [Which I am leaning somewhat more towards] or just 1 [Which if it is, can you tell me why]? And if not either, please just tell me the answer... :(

4. Feb 14, 2013

### tcc88

I am using 1/6.32 and 1/8.16 and I am still getting the wrong answer?!?! WTH is going on?!?

5. Feb 14, 2013

### cepheid

Staff Emeritus
The thing I was trying to hint at very strongly was that the 6.32 and 8.16 ARE the angular frequencies. I even said that they tell you how OFTEN or FREQUENTLY a revolution occurs.

The angular frequency is also the angular speed in this case. Do you know how to find the linear speed given the angular speed?

6. Feb 14, 2013

### tcc88

Would the Linear speed be rps[angular speed or frequency] * pi * d? Also will I need both to find the circular acceleration or just one? I legit don't remember my professor teaching this, but I am willing to learn it if it means getting my h.w in on time! :/

7. Feb 14, 2013

### cepheid

Staff Emeritus
No, the linear speed is the angular speed multiplied by r (the radius). The reason for this comes from the definition of an angle (in radians). The angle is the arc length you travel around the circle divided by the radius. Linear speed would be arc length/time, and if you divided this by the radius you'd get angle/time, which is angular speed. Make sense?

If you went a full revolution, the distance (arc length) would be 2*pi*r (a full circumference) and so the angle would be this divided by r, which would be 2pi radians.

8. Feb 14, 2013

### cepheid

Staff Emeritus
You have to be careful to convert the angular speed from rev/sec to radians/sec before working with these formulas. As I said above, one revolution is 2*pi radians.

9. Feb 14, 2013

### Andrew Mason

Just to follow up on what Cephid has said, it is easier and less confusing to use ac2r rather than converting to tangential speed. ω=2πf where f is the frequency of rotation = 1/period of rotation (f=1/T).

AM