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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.

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.

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?

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.

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?