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Homework Help: Stuck on a rotational kinematics question

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A barrel is lowered into a boat using the illustrated apparatus. The barrel can be considered to be a uniform cylinder with M=100kg and r=0.40m. The weight on the other end of the rope has m=30kg. Assume that the barrel does not slip against the wall, that the other 2 pulleys are massless and frictionless and the rope does not stretch and has no mass. What is the linear acceleration of the mass m? Answers are in m/s^2.

    2. Relevant equations
    3. The attempt at a solution
    So basically I set the tension in the rope as T, and did an Fnet equation for the small block. From there I got Fnet = T-30(9.8) = 30a
    So I isolated for T and got T=30(a+9.8)
    From that I used T to find the torque that it exerted on the cylinder. Now this is where I run into the problem, I know torque = T x R, but do I treat the point of contact between the cylinder and the wall as the pivot point, so R=0.8, or do I treat the center of the cylinder as the pivot point, so R=0.4?
    Also, can the center of gravity of the cylinder itself produce rotational torque about the point of contact between the wall and the cylinder (i.e. torque = mgr)?
  2. jcsd
  3. Dec 10, 2012 #2
    Good, your first part for mass m is correct.
    When observing the illustration, whereabout do you think the rotation takes place? about the centre as you ask or on the wall? Think of it as a block, if I have a block on a horizontal floor which I pull with a rope which is attached at the top right corner (2D image) whereabout is the rotation of the block? Right lower corner?
    You see where I am going with this. Draw a FBD for mass M now and use your equations as you stated...
  4. Dec 10, 2012 #3


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    Homework Helper

    In principle, you can write the torque equation both for the instantaneous axis and the central axis of the barrel. As you have the acceleration of the end of rope, which is the same as the linear acceleration of the outer rim of the barrel, it is easier to write the torque equation for the instantaneous axis. Of course, the weight of the barrel also produces torque.

  5. Dec 10, 2012 #4
    So basically for the FBD of the big barrel, I've got Fnet = Fg - (T+fs) where fs is friction. From this I get:
    fs = Fg - T = (100)(9.8) - 30(9.8 + a)
    Now I know torque = fs * radius, and that tension and friction work in opposite directions.
    I know it sipns counter clockwise, so:
    net torque = (fs * r) - (T * r) = (0.4)(980 - 294 - 30a) - (12)(9.8 + a)
    = 274.4 - 12a - 117.6 - 12a
    =158.8 - 24a
    We know net torque = I * (a/r), and I used the parallel axis theorem for the cylinder to get I.
    So (1/2)MR^2 + MR^2 which is (3/2)(MR^2) = 3/2(100 * 0.4^2) = 24.
    So net torque = 24 * (a/r) = 24 * (a/0.4) = 60a
    So 158.8-24a = 60a
    158.8 = 84a
    Therefore, a = 158.8/84 which is 1.87.
    Now, the question is did I do it right and should I just accept some sort of rounding error and take 1.8 as my answer, or did I do something wrong?
  6. Dec 10, 2012 #5
    Just realized I forgot that for the barrel, Fnet = 100 * a (and a is the same for both masses)
    So my new revised process is fs = Fg - Fnet - T
    = (100)*(9.8) - 100a - 30(9.8 + a)
    =980 - 294 - 100a - 30a
    = 686-130a
    so net torque = (fs * r) - (T * r)
    = 0.4(686-130a) - 0.4* 30(9.8 + a)
    = 156.8 - 52a - 12a is equal to I * (a/r)
    From above, 156.8 = (60 + 52 + 12)a = 124a
    So a = 156.8/124 = 1.265
    So clearly that approach didn't work...
  7. Dec 10, 2012 #6
    Now, using the instantaneous axis, I get that net torque = mg * r - (t * 2r)
    = 980 * 0.4 - 0.8(30)(9.8 + a)
    = 392 - 235.2 - 24a
    = 156.8 - 24a
    Which is again equal to I *(a/r) which is 24 * a/0.4 = 60a
    so 156.8 = 84a
    a = 156.8/84 = 1.86666667 again (is this the answer?)

    Also, on a quick side note, if I am treating the centre of mass as the point of rotation as I did above, does my moment of inertia change to that of just the cylinder? i.e. do i not have to use the parallel axis theorem anymore? If that's the case then I = (1/2)(MR^2) = 8 so net torque = 8/0.4 * a = 20a
    Using this new I while treating the centre of mass as the pivot point, it comes to 1.866667 again.....
  8. Dec 10, 2012 #7


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    The solution is more clear if you do it symbolically and plug in numerical data at the end.

    What axis do you use? If it is the instantaneous axis, a=2r α.

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