# Ap Physics Question.

by pezzang
Tags: physics
Mentor
P: 41,581
 Originally posted by pezzang PLEASE REFER TO THE PICTURE ATTACHED.
I don't see the picture.
 Mentor P: 41,581 Ap Physics Question. Use the attachment feature.
 P: 29 i tried but it does not work..ok i will explain instead in the first picture for question part (a) only, there are two regular circles. circle A has a radius of R and circle B has a radius 2R. They have the same mass and are indeendent of each other. in the second picture which will be used from part (b) to (e), the two circles described in the first picture are in a chain. the circle B on the right and Circle A on the left. the chain moves from left to right and the tension on the bottom part at first is zero. i hope this helps to solve the question. Thank you!
Mentor
P: 41,581
 Originally posted by pezzang -> I used the equation Inertia = mass*radius to solve this question, Let's say that smaller disk has a subscript "A" and larger; "B". Then, I(B) = m(2R) = mR(2). since mR is equal to I, we have I(B) = I*2 = 2I so the answer is 2*I. Am I right?
First things first: What's the moment of inertia of a disk about a central axis? It's I = 1/2 M R2

Redo the first part, then we'll continue.
Mentor
P: 41,581
Let me give you some tips on the other parts.

 Originally posted by pezzang (b) The angular acceleration of the larger disk (alpha(B)) -> torque(A) = I*alpha torque(B) = 2I*alpha(B) torque(A) = torque(B) <- IS IT POSSIBLE THAT I ASSUME THIS?????? I*alpha = 2I*alpha(B) alpha(B) = alpha / 2. <- AM I RIGHT?
Forget about torques for this part. &alpha;A is a given. What's the relationship between &alpha;A and &alpha;B? Hint: the two disks are tied by a chain. What does that mean?

Also, no, you cannot assume that the torque is the same on both disks! Think about it. DiskA has two forces acting on it: the force applied by the student in giving his torque, and the tension in the chain pulling the other way. DiskB has only a single force on it.
 (c) The tension in the upper part of the chain -> T(Tension) = (I*alpha) / R. T(A) = (I*alpha) / R T(B) = (2I)(alpha(B)) / (2R) = (2I*alpha/2) / R = (I*(alpha) / (2R) TO FIND TENSION, DO I SUBTRACT THE SMALLER ONE FROM LARGER ONE?? IF I DO, I GET: T(A) = T(B) = (I*alpha) / R - (I*alpha) / (2R) = (I*alpha) / (2R) Therefore, the tension in teh upper part of the chain equals to (I*(alpha) / (2R). (d) The torque that the student applied to the smaller disk -> torque(A) = I*alpha <- I found this equation in my physics book. Is right? I kind of doubt it though. Please help me.
The way to solve c & d is to write down the torque equations for both disks and solve the two equations together.
Torquenet=I&alpha;, for each disk. (You'll need the results from parts b and a.)
 (e) The rotational kinetic energy of the smaller disk as a function of time -> RKE = (1/2)*I*w^2 (omega) to find an equation of w(omega) in terms of t and alpha, we take antiderivative of the equation alpha = (dw)/(dt), then we get: alpha*t = w IF we substitute alpha*t for w(omega) in the equation RKE = (1/2)*I*w^2 (omega), we get RKE = (1/2)I*(alpha*t)^2. So that is the answer.
Yes!
 P: 29 Than you so much. Here is what I have so far after getting your advice. (a) Moment of Inertia of a disk is equal to I = (MR^2) / 2. So, I(B) = (1/2) (M)(2R)^2 = 2MR^2. Therefore, the answer is 2MR^2. (b) Since the disks are connected by a chain, they have the same angular acceleration. Thus the angular acceleration of the larger disk is equal to just メA, which is given in the problem. (c) I do not know how to find tension. I guess the tension on the upper part of the chain would equal to torque(B) - torque(a). Since torque is equal to I*メ, and メ is same for disk A and B, Tension = (1/2)*M(2R^2)*メ - (1/2)(MR^2)*メ = (3/2)MR^2*メ. So, the answer is equal to (3/2)MR^2*メ. (d) torque of the smaller disk is equal to (1/2)(MR^2)*メ, which we found in part(c). How do they look now? Much closer??? Thank you for your helpful suggestions. Please leave me more feedbacks~!
Mentor
P: 41,581
 Originally posted by pezzang Than you so much. Here is what I have so far after getting your advice. (a) Moment of Inertia of a disk is equal to I = (MR^2) / 2. So, I(B) = (1/2) (M)(2R)^2 = 2MR^2. Therefore, the answer is 2MR^2.
Yes. Now put it in terms of IA= I;
IB= 4*(1/2MR2)= 4I
 (b) Since the disks are connected by a chain, they have the same angular acceleration. Thus the angular acceleration of the larger disk is equal to just メA, which is given in the problem.
No, they have the same tangential acceleration, not angular acceleration. Think about it: the same length of chain must pass along each disk. Imagine the disk B was 1000 times bigger than disk A: the same bit of chain that turns A half way around would barely budge disk B.
 (c) I do not know how to find tension. I guess the tension on the upper part of the chain would equal to torque(B) - torque(a). Since torque is equal to I*メ, and メ is same for disk A and B, Tension = (1/2)*M(2R^2)*メ - (1/2)(MR^2)*メ = (3/2)MR^2*メ. So, the answer is equal to (3/2)MR^2*メ. (d) torque of the smaller disk is equal to (1/2)(MR^2)*メ, which we found in part(c).
Write out the torque equations for each disk. Tension is force; TensionXradius is the torque exerted by the tension.
 How do they look now? Much closer??? Thank you for your helpful suggestions. Please leave me more feedbacks~!
You're getting closer. Keep going!
 P: 29 Thank you again for your advice. I have found part (b) but i really can't solve part (c) and (d). Can you please explain the answer in detail? Meanwhile here is part (b) (b) a(tan) = (dv)/(dt) = r*(dw)/(dt) = r*alpha. a(tanA) = a(tanB) R*alpha = (2R)*alpha(B) alpha(b) = (R*alpha)/(2R) = alpha / R. thus, the answer is alpha / R. PLEASE TELL ME PART (C) AND (D). I AM REALLY REALLY REALLY CURIOUS~!!!
Mentor
P: 41,581
 Originally posted by pezzang Meanwhile here is part (b) (b) a(tan) = (dv)/(dt) = r*(dw)/(dt) = r*alpha. a(tanA) = a(tanB) R*alpha = (2R)*alpha(B) alpha(b) = (R*alpha)/(2R) = alpha / R. thus, the answer is alpha / R. PLEASE TELL ME PART (C) AND (D). I AM REALLY REALLY REALLY CURIOUS~!!!
You messed up the algebra above: &alpha;B= &alpha;/2

OK, OK, you've suffered enough...

The torque equation for Disk B:
FTRB = IB &alpha;B
Which, when we use the results of a & b, becomes:
FT = I&alpha;/R(That's the answer to part C!)

The torque equation for Disk A:
&Tau;-FTR = I &alpha; (where FT is the tension; &Tau; is the the applied torque)

Now we can solve for &Tau; (using the answer from part C)
&Tau; = 2I&alpha; (That's the answer to part D.)

Make sense? Go over this until you understand it. (And you'd better check my algebra.)
 P: 29 I TRULY APPRECIATE YOUR HELP ALONG THE WAY. EVERYONE OF US WHO GETS HELP FROM YOU ARE GRATEFUL TO YOU!!!! THANK YOU SO SO SO SO SO MUCH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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