Linear and angular acceleration of a falling cylinder? [long]by Bob Loblaw Tags: acceleration, angular, cylinder, falling, linear 

#1
Oct3107, 01:21 AM

P: 68

1. The problem statement, all variables and given/known data
http://img248.imageshack.us/img248/1639/cylinderjl0.jpg A uniform cylinder with a radius of R and mass M has been attached to two cords and the cords are wound around it and hung from the ceiling. The cylinder is released from rest and the cords unwind as the cylinder descends. (a) draw a proper free body diagram for the cylinder; (b) Apply Newton’s second law to the cylinder; (c) apply Newton’s second law in rotational form to the cylinder; (d) the two equations you have written so far contain three unknowns; what is the relationship between the linear acceleration of the cylinder and its angular acceleration? (e) Solve for the linear acceleration of the cylinder; (f) What is the tension in the cords? The attempt at a solution A) The free body diagram should include the forces of tension pulling the cylinder up and the weight of the cylinder, right? B) Essentially it would be Tensionweight of cylinder=angular acceleration, is that correct? C)I know that Newton's second law in rotational form is: Net external torque = moment of inertia x angular acceleration So I would need to solve for moment of inertia: I=mr^2 X Angular acceleration = net external torque. D) So I have Tensionweight of cylinder=angular acceleration I=mr^2 X Angular acceleration = net external torque. Linear Acceleration and Angular Acceleration are related by: a. The Mass. b. The Radius. c. The Torque. d. The Force. In this problem, the mass, radius, and torque are unknown. The force working on the cylinder I believe is just gravity. E and F) I am not sure how to do these. I know this is a lengthy problem, but any help of guidance would be appreciately greatly! Even if ou don't know  I would settle for just a word of encouragement. 



#2
Oct3107, 01:46 AM

HW Helper
P: 4,125

so write the [tex]\Sigma F_y = ma[/tex] equation... net torque = I*alpha (alpha is angular acceleration) what goes in the left side of this equation? 



#3
Oct3107, 01:59 AM

P: 68

Thanks. I made some changes and now I have:
A) My free body diagram has two tension forces up and weight pointed down. B) Newton's 2L looks like 2TW=ma C) net torque = I*alpha I=1/2mv^2 for a cylinder so: net torque = 1/2mv^2*alpha. I am not certain about the left side. Is net torque equal to zero? I am not sure of the relationship between linear and angular velocity. My guess is that the radius of the cylinder will roll as it is unwound from the rope in such a way that the rotations per second will relate to the speed of the fall? 



#4
Oct3107, 02:06 AM

HW Helper
P: 4,125

Linear and angular acceleration of a falling cylinder? [long]net torque = 1/2mr^2*alpha. (not v but r). v = rw that means that a = r*alpha 



#5
Oct3107, 02:12 AM

P: 68

By definition, torque=force x radius. The relationship between tension and torque isn't clear to me right now.




#6
Oct3107, 02:26 AM

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P: 4,125

force*(perpendicular distance from the axis to the line of the force) The force is T. Think of the axis of the cylinder... and the line of the force... draw a line segment going from the axis... to the line of the force(this line segment is just a radius of the cylinder)... the length of that line segment is r... So torque due to one rope is T*r. Torque due to both ropes is 2T*r. 



#7
Oct3107, 02:38 AM

P: 68

Thanks, I see the problem a little better now.
So: Newtons 2L: 2TW=ma Rotational form: 2T*r=1/2mr^2*alpha The relationship between the two forms: a = r*alpha So we want to solve for linear acceleration. Would we take r as (1/2mr^2*alpha)/2T? If so, that would still leave me to deal with another unknown, alpha. Thanks for your help so far. I feel fortunate that you were able to assist me. 



#8
Oct3107, 02:46 AM

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P: 4,125

2T*r=1/2mr^2*alpha leaving 2T=1/2mr*alpha so you have the 3 equations: 2Tmg=ma 2T=1/2mr*alpha a = r*alpha trying playing with these 3 equations... you should be able to solve for a, alpha and T in terms of m, g and r... try to go from 3 equations in 3 unknowns to 2 equations in 2 unknowns... 



#9
Oct3107, 02:52 AM

P: 68

I came up with 2r*alpha=a and mr*alpha=T
Is it even possible to have nonsymbolic answers for this problem? 



#10
Oct3107, 02:58 AM

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P: 4,125

I blundered... for the first equation we should use: mg  2T = ma (1)... since the object is going downwards... 2T=1/2mr*alpha (2) a = r*alpha (3) plug in a from (3) into (1). that gives: mg  2T = mr*alpha ... solve this along with (2): 2T=1/2mr*alpha solve the above 2 equations... you should be able to get T and alpha in terms of m, r and g... 



#11
Oct3107, 03:08 AM

P: 68

Solving for T in equation 2 gives me:
T=mr*alpha Putting that into the equation derived from plugging (3) into (1) mg2mr*alpha=mr*alpha that leaves me with mg=3 or m9.81=3 or .31kg Am I on the right track? The problem asks for linear acceleration of the cylinder as well as tension though I am not sure if I am any closer to finding a real solution. EDIT: Fixed bad equation and got new results. 



#12
Oct3107, 03:17 AM

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P: 4,125

equation 2 is: 2T = 1/2mr*alpha gives T = 1/4 mr*alpha plugging (3) into (1) (this is the equation I changed) gives: mg  2T = mr*alpha then plugging in the T gives: mg  2(1/4 mr*alpha) = mr*alpha we can cancel m's g  (1/2)r*alpha = r*alpha gives (3/2)r*alpha = g solving for alpha gives alpha = 2g/(3r) plugging that into T = 1/4 mr*alpha gives T = (1/4)mr*(2g)/(3r) so T= (1/6)mg and a = r*alpha = r*2g/(3r) so a= 2g/3 check over this yourself... I could have blundered... look over my last post with the 3 equations again... try going through solving it and see if you get the same answers... 



#13
Oct3107, 03:28 AM

P: 68

I blundered on my algebra as well. I should have:
mg2T=mr*alpha T=1/4mr*alpha mg2(1/4)mr*alpha=mr*alpha cancelling the mr*alpha leaves me: mg=1/2 9.81m=1/2 m=.05kg 



#14
Oct3107, 03:31 AM

P: 68

I came out with the same answers (once I knew what to do)
so in real numbers a=2(9.81)/3=6.54 T=(1/6)mg=(1/6)(1/2)=1/12N Is this beast finally slain? 



#15
Oct3107, 03:32 AM

HW Helper
P: 4,125

mg2(1/4)mr*alpha=mr*alpha if you divide each term by mr*alpha, you get mg/(mr*alpha) 1/2 = 1 g/(r*alpha)  1/2 = 1 



#16
Oct3107, 03:33 AM

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P: 4,125





#17
Oct3107, 03:36 AM

P: 68

You're right about my algebra being bad. So now I cannot see how mass would be solvable. If that is the case I won't get a real numerical answer. Is that normal for these types of problems?




#18
Oct3107, 03:40 AM

HW Helper
P: 4,125

so: alpha = 2g/(3R) T= (1/6)Mg so a= 2g/3 is perfectly acceptable as answers... EDIT: If they give numbers for M and R, then of course, we'd need to get numerical answers for alpha, T and a... but without numbers, this is just fine. 


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