# Homework Help: Circular motion

1. Mar 29, 2005

### john_f2004

An exhaust fan rotates at 9.0 x 10^2 revolutions per minute. Find the acceleration of a point on a blade at a distance of 25cm from the axis of rotation.

I don't know where to begin. Can someone help?

2. Mar 29, 2005

### dextercioby

How about:what theory would have to use to get it solved...?Formulas and a reasoning for using them...

Daniel.

3. Mar 29, 2005

### john_f2004

Ok I think the ? has to do with uniform circular motion. The only equation I am given is a (sub) c = v^2 divided by r
A (sub) c = centripetal acceleration
v = velocity

4. Mar 29, 2005

### dextercioby

$a_{cp}=\omega^{2} r$ is the formula that u need.

U have been give the frequency.Find the angular velocity "omega"...

Daniel.

5. Mar 29, 2005

### john_f2004

isnt that the same formula i just gave you?

6. Mar 29, 2005

### dextercioby

Not really.There's a close connection between the 2,though.So do it.

Daniel.

7. Mar 29, 2005

### john_f2004

that is the problem I don't know how

8. Mar 29, 2005

### dextercioby

$$a_{cp}=(2\pi\nu)^{2}r$$

U know "r" and u know "nu",now find the acceleration.Convert from revs/min to Hz...

Daniel.

9. Mar 29, 2005

### john_f2004

the question has nothing to do with frequency

10. Mar 29, 2005

### dextercioby

If you don't see it,that's bad.

"rotates at 9.0 x 10^2 revolutions per minute"...what does that mean?

Daniel.

11. Mar 29, 2005

### john_f2004

the answer i am supposed to get has nothing to do with frequency. all i need to know is the acceleration. no where in our notes and in our class did we talk about frequency.

12. Mar 30, 2005

### dextercioby

Incidentally,in this problem you are given the frequency & the distance between the point & the center of the circle.

I can't do anything more.I've already told u what to do...

Daniel.

13. Mar 30, 2005

### tutor69

Johnf2004,
In the given quetion you are given the frequency,f i.e. number of revolutions per minute. You can convert this into time period, T using:
T = 2*pi/f

Now use $$v = \frac{2 \pi r}{T}$$
as for an object moving with constant speed on a circular path, v = s/t
Here, the distance,s is the circumference(2*pi*r) traversed in time t. But then t, the time is also equal to time taken for one complete revolution, hence equal to T.

acceleration = (v^2)/r

What dextercioby has explained is same , he started with $$\omega$$, the angular velocity

Last edited: Mar 30, 2005