
#1
Nov2103, 06:59 AM

P: 29

Hi, thank you so much for making this kind of post for students like me. I have tried the quesotins but i want to make sure that i was right. if worng, please correct me.
THANK YOU SO MUCH. Also i had attached pictures of the question to give you enough information. PLEASE REFER TO THE PICTURE ATTACHED. Q) The two uniform disks shown above have equal mass, and each can rotate on frictionless bearings about a fixed axis through its center. The smaller disk has a radius R and moment of inertia I about its axis. (a) determine the moment of inertia of the larger disk about tis axis in terms of I. > 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? The two disks are then linked as shown below by a light chain that cannot slip. They are at rest when, at time t = 0, a student applies a torque to the smaller disk, and it rotates counterclockwise with constant angular acceleration alpha. Assume that the mass of the chain and the tension in teh lower part of the chain are negligible. In terms of I, R, alpha and t, determine each of the following. (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? (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. (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. So please correct me and thank you so much. Have a wonderful day, people!!!!!! 



#3
Nov2103, 09:59 AM

P: 29

can i send you email so that you can see the pictures???
Please give me your email address. Thank you, again. 



#5
Nov2103, 12:18 PM

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! 



#6
Nov2103, 12:37 PM

Mentor
P: 40,878

Redo the first part, then we'll continue. 



#7
Nov2103, 01:06 PM

Mentor
P: 40,878

Let me give you some tips on the other parts.
Also, no, you cannot assume that the torque is the same on both disks! Think about it. Disk_{A} 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. Disk_{B} has only a single force on it. Torque_{net}=Iα, for each disk. (You'll need the results from parts b and a.) 



#8
Nov2103, 02:32 PM

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~! 



#9
Nov2103, 02:51 PM

Mentor
P: 40,878

I_{B}= 4*(1/2MR^{2})= 4I 



#10
Nov2103, 05:48 PM

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~!!! 



#11
Nov2203, 05:26 AM

Mentor
P: 40,878

OK, OK, you've suffered enough... The torque equation for Disk B: F_{T}R_{B} = I_{B} α_{B} Which, when we use the results of a & b, becomes: F_{T} = Iα/R(That's the answer to part C!) The torque equation for Disk A: ΤF_{T}R = I α (where F_{T} is the tension; Τ is the the applied torque) Now we can solve for Τ (using the answer from part C) Τ = 2Iα (That's the answer to part D.) Make sense? Go over this until you understand it. (And you'd better check my algebra.) 



#12
Nov2203, 08:35 AM

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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