- #1

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

- Thread starter john_f2004
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- #1

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

- #2

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Open your book.

Daniel.

- #3

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A (sub) c = centripetal acceleration

v = velocity

r = radius

- #4

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

Daniel.

- #5

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isnt that the same formula i just gave you?

- #6

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

Daniel.

Daniel.

- #7

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that is the problem I don't know how

- #8

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

Daniel.

- #9

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the question has nothing to do with frequency

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

Daniel.

- #11

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- #12

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

Daniel.

- #13

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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 [tex] v = \frac{2 \pi r}{T} [/tex]

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 [tex]\omega[/tex], the angular velocity

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 [tex] v = \frac{2 \pi r}{T} [/tex]

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 [tex]\omega[/tex], the angular velocity

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