Find the Acceleration of a Point on an Exhaust Fan Blade at 25cm from Axis

AI Thread Summary
To find the acceleration of a point on an exhaust fan blade at 25 cm from the axis of rotation, the discussion emphasizes the use of centripetal acceleration formulas. The relevant equations include a_c = v^2/r and a_cp = ω^2 r, where v is the linear velocity and r is the radius. Participants highlight the need to convert the fan's rotation speed from revolutions per minute to frequency in Hertz to use these formulas effectively. The connection between angular velocity and frequency is clarified, indicating that frequency is crucial for determining the time period and subsequently the linear velocity. Ultimately, the acceleration can be calculated using the provided values and the derived formulas.
john_f2004
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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?
 
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How about:what theory would have to use to get it solved...?Formulas and a reasoning for using them...

Open your book.

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

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

Daniel.
 
isnt that the same formula i just gave you?
 
Not really.There's a close connection between the 2,though.So do it.

Daniel.
 
that is the problem I don't know how
 
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.
 
the question has nothing to do with frequency
 
  • #10
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
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
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
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
 
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