Uniform Circular Motion, Acceleration problem

AI Thread Summary
The discussion centers on a physics problem involving a cat on a merry-go-round exhibiting uniform circular motion. The cat's velocities at two different times are given, and the first part of the problem requires calculating the centripetal acceleration, which is determined to be 5.2 m/s². For the average acceleration, there is confusion regarding whether to consider the magnitudes of the velocities or their vector forms. It is clarified that the velocities at the two times are indeed opposite, indicating a complete revolution, and the average acceleration should be expressed in unit vector notation. The conversation also touches on formatting equations and vectors in the forum.
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Homework Statement


A cat rides a merry-go-round turning with uniform circular motion. At time t1 = 2.00 s, the cat’s velocity is V1 = (3.00 m/s)i + (4.00 m/s)j , measured on a horizontal xy coordinate system. At t2 = 5.00 s, the cat’s velocity is V2 = (3.00 m/s)i + (4.00 m/s)j.
What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s average acceleration during the time interval t2 - t1, which is less than one period?

Homework Equations


[/B]
T = 2πr/V
a = V2/r

The Attempt at a Solution


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So, the first thing I did was sketch the situation, and the vectors seem to be on the opposite end of the circumference, so i tested for that:

V1 * V2 = V1x * V2x + V1y * V2y = 3*(-3) + 4*(-4) = -25
V1 * V2 = V1 * V2 CosΘ = √32+42*√(-3)2+(-4)2 = 25 CosΘ → CosΘ = -1 → Θ = Cos-1(-1) = 180°

So it's proven that the cat is in two different positions that cover half of the circumference, which means that
Δt = ½T
Δt = t2-t1 = 5s-2s = 3s
T = 2Δt = 3s * 2 = 6s
T = 2πr/V → r = TV/2π = 6s * 5 m/s / 2π = 4,8 m
a = V2/r → (5 m/s)2/4,8 m = 5,2 m/s2

So this should answer the first part of the problem, now I'm asked to find the average acceleration, and I'm not sure what approach is the correct one, I know the equation for the average acceleration is:

aavg = V2 - V1 / t2 - t1

And I am not sure if I should consider just the magnitude of the two velocities, which means that aavg = 0, or if I should solve in unit vector notation, doing the following:

aavg = V2 - V1 / t2 - t1 = (-3-3)i + (-4-4)j / 3s = -2i -2,67j → aavg = √(-2)2+(-2,67)2 = 3,3 m/s2

What is the correct way to answer the second question?

(also, kinda off topic, is there a way to write fractions and vectors on the forum?)

Thanks in advance, Ivan.
 
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Umh, the spam filter is not letting me edit the post, but of course the equation for the average acceleration is

aavg = V2 - V1 / t2 - t1

And I have used that in the calculations.
 
No1_129848 said:
At time t1 = 2.00 s, the cat’s velocity is V1 = (3.00 m/s)i + (4.00 m/s)j , measured on a horizontal xy coordinate system. At t2 = 5.00 s, the cat’s velocity is V2 = (3.00 m/s)i + (4.00 m/s)j.
Is this correct? It looks like V1 = V2 which means that the cat has made a complete revolution. Did you miss a negative sign somewhere?
 
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Oh, indeed, both signs are negative in V2, i can't edit the OP, but the correct data is
V1 = (3.00 m/s)i + (4.00 m/s)j
V2 = (-3.00 m/s)i + (-4.00 m/s)j
 
Part (a) looks fine.
You must use unit vector notation to find the angular acceleration. I would also say that, since the question asks for the average acceleration and not its magnitude, you should leave in unit vector form.
Yes, there is a way to write fractions and all sorts of other algebraic expressions. Click on the link "LaTeX", near bottom left next to the question mark.
 
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Thanks for the help, much appreciated!
 
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