# Centripetal Acceleration HELP. Is it challenging? Well come find out!

1. Oct 22, 2013

### TwoTeez

1. An athlete whirls a 7.48 kg hammer tied to
the end of a 1.4 m chain in a horizontal circle.
The hammer moves at the rate of 1.76 rev/s.
What is the centripetal acceleration of the
hammer? Assume his arm length is included
in the length given for the chain.

2. Equations I am pretty sure I need are
v = 2∏r/T
and
ac= v2/r
not sure if I need any others.

3. I tried finding v first, and I assumed the 1.76 rev/s was T, so I did 2*∏*1.4(about 8.80) and divided it by 1.76, getting about 4.99 m/s
and I plugged it in to the ac equation.
So I squared the 4.99, getting about 24.98, and dividng it by 1.4, which gets me about 17.843.
I don't know what the answer is.

I feel like what I am doing wrong is the 1.76 rev/s is not T, or it is, but I have to change it so it's in like m/s or something like that. If that's the case, then I need help finding that because I have no idea how to do that.

Thank you for any help!

2. Oct 23, 2013

### serllus reuel

in your equation for v, T is the time to complete one revolution (make sure you see why)

1.76 rev/s

is that what you want? or is it seconds per rev?

3. Oct 23, 2013

### TwoTeez

I think I see where you are getting at.
For every second that passes, it makes 1.76 revolutions.
But to find the velocity, I need to put T as a time per revolution. So would I just do the inverse of T?
So if i do the inverse for T, wouldn't that just be multiplying T with the 2∏r?
But wouldn't I need to the the inverse of everything else as well? What you do to one side of the equation, you do to the other.

So 1/v = T/2∏r
And then do the inverse again to get the answer. But isn't that the same is the answer I would have gotten before? Am I even going in the right direction?

4. Oct 23, 2013

### serllus reuel

well 1.76 is not T. Your messing around with thinking that it is. 1.76 is the angular speed, completely different from T.

the equation v = 2∏r/T is correct. Find T.

5. Oct 23, 2013

### TwoTeez

If I want to turn the rev/s into common terms, would I multiply 1,76 by 2∏r?

6. Oct 23, 2013

### TwoTeez

I have yet to learn about angular speed in my class, so I know practically nothing on it.

7. Oct 23, 2013

### twj2944

With what you have now, you have 1.76 revolutions over one second. Just do the inverse of that: (1/1.76) to find your period, T

This is the relationship between frequency and period.

You are given a frequency of 1.76 rev/sec.

Frequency is (1/T) T is the period