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Homework Help: Linear and angular acceleration of a falling cylinder? [long]

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img248.imageshack.us/img248/1639/cylinderjl0.jpg [Broken]

    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 Tension-weight 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
    Tension-weight 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.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 31, 2007 #2


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    not angular acceleration. The linear acceleration of the center of mass...

    so write the [tex]\Sigma F_y = ma[/tex] equation...

    Yes, but the moment of inertia of the cylinder isn't mr^2... look that up... so:

    net torque = I*alpha (alpha is angular acceleration)

    what goes in the left side of this equation?

    So in this part you need to relate a and alpha... do you know the relationship between velocity and angular velocity?

    Once you write your equations in parts B, C, and D... just solve them to get the answers to E and F.
    Last edited by a moderator: May 3, 2017
  4. Oct 31, 2007 #3
    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 2T-W=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?
  5. Oct 31, 2007 #4


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    yes, exactly.

    you mean:

    net torque = 1/2mr^2*alpha. (not v but r).

    What is the definition of torque? How do the tensions in the cords relate to the torque?

    if an object rolls without slipping... like a tire on a road... then:

    v = rw

    that means that

    a = r*alpha
  6. Oct 31, 2007 #5
    By definition, torque=force x radius. The relationship between tension and torque isn't clear to me right now.
  7. Oct 31, 2007 #6


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    yes, that's the definition. the way I think of it is... torque is

    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.
  8. Oct 31, 2007 #7
    Thanks, I see the problem a little better now.

    Newtons 2L:
    Rotational form:

    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.
  9. Oct 31, 2007 #8


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    You can cancel 1 r in:




    so you have the 3 equations:

    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...
  10. Oct 31, 2007 #9
    I came up with 2r*alpha=a and mr*alpha=T

    Is it even possible to have non-symbolic answers for this problem?
  11. Oct 31, 2007 #10


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    No. you'll have to give your answer in terms of r, m and g. But your equations don't look right...

    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):

    solve the above 2 equations... you should be able to get T and alpha in terms of m, r and g...
    Last edited: Oct 31, 2007
  12. Oct 31, 2007 #11
    Solving for T in equation 2 gives me:
    Putting that into the equation derived from plugging (3) into (1)

    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.
    Last edited: Oct 31, 2007
  13. Oct 31, 2007 #12


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    I messed up on the first equation... I modified my previous post now...

    equation 2 is:

    2T = 1/2mr*alpha
    T = 1/4 mr*alpha

    plugging (3) into (1) (this is the equation I changed)


    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...
  14. Oct 31, 2007 #13
    I blundered on my algebra as well. I should have:


    cancelling the mr*alpha leaves me:
  15. Oct 31, 2007 #14
    I came out with the same answers (once I knew what to do)

    so in real numbers a=2(9.81)/3=6.54

    Is this beast finally slain?
  16. Oct 31, 2007 #15


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    you can't cancel mr*alpha (because of the mg)... we have to divide all the terms by the same thing:


    if you divide each term by mr*alpha, you get

    mg/(mr*alpha) -1/2 = 1
    g/(r*alpha) - 1/2 = 1
  17. Oct 31, 2007 #16


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    the mg = 1/2 isn't right.
  18. Oct 31, 2007 #17
    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?
  19. Oct 31, 2007 #18


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    Yeah, it is definitely normal. The idea is to solve in terms of the variables given... you're given a mass of M and radius R (we used small letters but no matter)... so your answers have to be in terms of those variables... and using constants like g is fine.


    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.
    Last edited: Oct 31, 2007
  20. Oct 31, 2007 #19
    I really appreciate your help.
  21. Oct 31, 2007 #20


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    no prob. :)
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