Uniform Circular Motion: Centripital Acceleration vs. Acceleration


by Chele
Tags: acceleration, centripital, circular, motion, uniform
Chele
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#1
Jun16-07, 05:39 AM
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I am very new to physics and am taking a my first University Physics class. This is not a call for a problem to be solved, but a clarification on terminology.

In solving problems for uniform circular motion, some problems call for the acceleration of the object (a=v^2/r) and others the centripital or instantanious acceleration (a=4pi^2r/T^2).

Can you please attempt to explain, in layman's terms, the difference between the two references to acceleration?

Thanks for your assistance.
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Jun16-07, 05:49 AM
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Quote Quote by Chele View Post
In solving problems for uniform circular motion, some problems call for the acceleration of the object (a=v^2/r) and others the centripital or instantanious acceleration (a=4pi^2r/T^2).
The two formulas are equivalent. (Express the speed in terms of circumference over period and you'll see for yourself.)

For uniform circular motion, the acceleration is centripetal. (Centripetal just means "towards the center".)
Chele
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Jun16-07, 09:58 PM
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Wow- I'll need to look at that in further detail....
Thanks for your help!

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Jun16-07, 10:57 PM
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Uniform Circular Motion: Centripital Acceleration vs. Acceleration


Quote Quote by Chele View Post
Wow- I'll need to look at that in further detail....
Thanks for your help!
Indeed, the two are equivalent for UCM. This can be seen easily if you recall that for constant speed, you may use v= distance/time. If you wait for the particle to go through a full circle, it will have covered a distance 2 Pi r, and the time elapsed will be the period T.
So

[itex] v_{ucm} = \frac{2 \pi r}{T} [/itex]

Using this formula it is simple to prove that the two equations for acceleration you gave are equal.
Chele
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#5
Jun17-07, 01:56 AM
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Okay, thanks. I worked it out and it is exactly the same. Not sure why I didn't see it before. Thanks guys!


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