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chandler742

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You are looking over a merry go round, and you notice that at t=0 it is rotating at 40 radians/sec. You decide to take two more readings, at t=1 sec (-10 rad/sec) and at t= 10 sec (-460 rad/sec).

Is the merry go round undergoing angular acceleration?

A) yes. the angular acceleration is counterclockwise.
B) yes. the angular acceleration is clockwise.
C) no. there is no angular acceleration
D) alright I give up.
 

wgu

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Originally posted by chandler742
You are looking over a merry go round, and you notice that at t=0 it is rotating at 40 radians/sec. You decide to take two more readings, at t=1 sec (-10 rad/sec) and at t= 10 sec (-460 rad/sec).

Is the merry go round undergoing angular acceleration?

A) yes. the angular acceleration is counterclockwise.
B) yes. the angular acceleration is clockwise.
C) no. there is no angular acceleration
D) alright I give up.

Assuming that you're not going to get a curve that is crazy...

it is going more negative in direction. 1st second goes -50rad/sec. next 9 second would go -450rad/sec and get you -460rad/sec if there's no acceleration, and indeed it is! Well that makes my clockwise q irrelevant. If it was I would probably have compared what it was @ 0 acceleration to actual and determine +/- acceleration.
 

freakazoid

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I would have to say it is undergoing acceleration, but I've got this bad feeling in my stomach that it's not acclerating at all. What's the convention for +/- in noting clockwise/counterclockwise?
 
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chandler742

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Here is a clue. Rotational motion problems are analogus to translational motion(i.e. you can use rotational equivalents of the equations of motion).


for example vf=vi + at, vf2=vi2+2ad,



The answer is C. at t=0, it is moving counterclockwise at 40
rads/sec
at t=1 it is moving counter clockwise at 10
rads/sec(i.e difference of 50 rad/s).
at t=10 it is still moving at 50 rad/s
500/10 sec.
Since, the angular velocity is not changing, there is no acceleration.


Keep in mind, you are responsible for rotational motion, it has appeared in a free response question on AAMC VI.

Two take home messages

1. + is counterclockwise, - clockwise
2. rotational motion problems are just like linear motion problems
(i.e. vf=vi+at--->wf=wi+alpha t) w=ang.velocity
alpha=ang. accerlation.

ON THE MCAT, angular acceleration will be constant unless you have to figure out the angular acceleration like the problem above.

good luck
 

freakazoid

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I'm still not getting it, chandler. It would make more sense to me that each of the values that you give us (40,-10,-460) are all analogous to velocity because it is giving us (change in angle)/(change in time).

Let me plug your values in vf=vi + at:

(-10 rad/sec) = (40 rad/sec) + a(1 sec)

Solving for a, we get a=-50 rad/s^2

Maybe in the question you meant the numbers (40, -10, -460) to be a quantity of displacement rather than velocity . . . but you can't tag rad/sec onto a value of displacement, can you?

Let me know.
 

chandler742

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hey freak,

you are right. I answered the question yesterday morning around 6 before coming into my lab, so I was not my usual analytical self.

The right answer is B. It has a angular acceleration counterclockwise.

Good job on the recovery.
 
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