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Seriously stumped by rotational motion

  1. May 14, 2003 #1
    Here's the problem I'm stuck on:

    An 8 lb. block slides on a horizontal table; between the block and table, coeffecient of friction = 0.25. The block is pulled by a string over a light, frictionless pulley wrapped around a solid cylinder of radius 2 ft. and weight 8 lb. The cylinder is released from rest. Compute the acceleration of the block and the tension of the string.

    The falling cylinder is basically like a yo-yo.

    The tension on both sides of the pulley are the same

    I've taken the torque about the center of the cylinder:
    Tension in the string = (mass cylinder)(tangential acceleration)/2

    I've also used Newton's Second Law with respect to the forces in the horizontal direction for the block and vertical direction for the cylinder.

    What I'm having trouble with is the relationship between the different accelerations in these equations.

    The cylinder is falling and unrolling. The acceleration that's referred to when you use NSL on the falling cylinder is the acceleration of the point where the string is unwinding, right? But how is this related to anything?

    Any hints would be greatly appreciated.

  2. jcsd
  3. May 14, 2003 #2
    I don't think so.
    I haven't tried to solve it yet so I'm not sure, but for the translational motion of the cylinder I think we have to use the vertical acceleration of the center of mass.

    ...and maybe figure the ANGULAR acceleration of the cylinder about the point where the string is unwinding...

    ...aha! I think I have it. Let me know what you come up with.
    Last edited: May 14, 2003
  4. May 14, 2003 #3
    Until I figure out how to relate the three different accelerations I've come up with, I can't find the solutions. And I can't for the life of me figure out how they're related. All the problems that I've been assigned prior to this one contained the condition that the rotating object did not slip, so the acceleration of the center of mass and the tangential acceleration were the same. This problem doesn't and I don't know how to proceed.
  5. May 14, 2003 #4

    Tom Mattson

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    They are related through the string. The "no slip" condition in this problem is that the string does not slip over the pulley. That means that the acceleration a of the string (which is the same as that of the block) is equal to Rα, where R is the radius of the pulley and α is its angular acceleration.
  6. May 14, 2003 #5
    You just have to look at it from a different perspective. Draw the string as a perfectly vertical line, tangent to the cylinder at "3 o'clock". Now, think of the cylinder as rotating about an axis located at that same point on its circumference where the string touches it. Of course, it can't really do that since that physical point is actually always rotating. But there always is some point at that relative location, and at any given instant, the cylinder can be thought of as rotating about that point. That means, then, that you can think of the center of the cylinder as rotating about that point on the circumference. Draw yourself a sketch of that situation & see if it helps you discover the amount of tension. (Be careful when you calculate the moment of inertia.)

    I didn't have a chance to check this approach with my physics prof., but it's giving me a very plausible-looking result so I'm pretty sure it's right.
  7. May 15, 2003 #6
    The pulley is light and frictionless and I'm not given any information about its radius, meaning that it shouldn't be considered in the problem.

    By the way the acceleration of the block is supposed to be 2ft/sec/sec.
  8. May 15, 2003 #7
    I AM treating the pulley that way (ie: ignoring it).

    I like this problem a lot. Seems like a good review for my final. But I don't agree with your result. I'm getting 2.67 ft/sec^2 for the block, and I'm still convinced that everything fits together perfectly, and I can't see anything wrong in my figures, so I guess it's time to show how I'm getting it. So if anybody sees a mistake in my approach, please straighten me out.

    cm = center of mass of the cylinder
    T = tension
    F = force (in general)
    f = force due to friction
    n = normal force
    T = torque
    A[b] = acceleration of block
    A[c] = linear (vertical) acceleration of cylinder
    A[t] = tangential acceleration of point on circumference of cylinder
    a = angular acceleration of cylinder
    r = radius of cylinder
    I[cm] = moment of inertia about center of mass of cylinder
    I[p] = moment of inertia about point on circumference of cylinder
    MG = weight

    OK: Now just focus on the cylinder. It is subjected to 2 forces, T acting vertically upward at a point p on the circumference, and MG, acting vertically downward at cm (center of mass). Since those are the only forces acting on it, it will experience linear acceleration (vertically downward) determined by:
    A[c] = F/m where F = T - MG, and m = MG/G = 8/32 = .25

    Also, it there will be angular acceleration because the two forces act at different points. So I choose my axis as point p where the cord is unwinding from the cylinder (the cord forms a tangent to the cylinder here). T acts at point p, so T exerts NO torque about my axis. The only torque is from MG, with a moment arm equal to r = 2 ft.
    Now, I[cm] of the cylinder is given by
    I[cm] = .5 mr^2 = .5*.25*2^2 = .5
    But that is not the axis of rotation. But by the parallel axis theorem:
    I[p] = I[cm] + md^2 where d=r so
    I[p] = .5 + .25*2^2 = 1.5
    a = T/I[p] = (8*2)/1.5 = 10.67
    Next, imagine the cm of the cylinder (instantaneously) tracing a circular path around point p. Then the linear (vertical) acceleration of the cylinder is equal to the tangential acceleration of cm,
    A[c] = r*a = 2*10.67 = 21.33 ft/s^2
    Let's give this a minus sign since it's direction is downwards: -21.33
    This lets us compute T, using F = mA on the cylinder:
    T + MG = mA
    T - 8 = mA
    T = 8 + mA = 8 + .25*(-21.33) = 2.67
    This works. If T=2.67 lb, there is a net downward force of 5.33 lb on the cylinder, which corresponds to vertical acceleration of -21.33 ft/s^2.

    Now, for the block, n = MG so
    f = .25*8 = 2 lb
    And if T=2.67 lb, the net horizontal force on the block is 2.67 - 2 = 0.67 lb so
    A = F/m = .67/.25 = 2.67 ft/s^2
  9. May 15, 2003 #8
    According to the textbook the Tension is 2.5 lb.

    I read your approach, but I don't understand how you came to your conclusions about the behaviour of the cylinder.

    ie. 1) Why did you use the parallel axis theorem. I see the cylinder moving translationally as it spins around its center of mass, so how would the P.A.T. apply? 2) Why would I imagine the cm of the cylinder tracing a circular path around point p?

    Your results are very close to the ones in the back of the text, so you might be on the right track. I'm not seeing why though.

    If you're correct, how do I acquire such vision?:smile:
  10. May 15, 2003 #9
    i dont understand how you americans can work with your imperial system. i guess it's just because i wasnt raised on it.
  11. May 15, 2003 #10
    I used the parallel axis theorem because I wanted to use that point p on the circumference as my axis of rotation -- just for convenience. We ARE allowed to choose any axis, correct?

    And from that, it follows that the cm is rotating about that point p.

    Anyway, if we now look at it as rotating about the cm, my figures still work.
    Let a[cm] = angular accel. about axis at cm

    I[cm] = .5*.25*2^2 = .5
    T[cm] = T * 2 = 2.67 * 2 = 5.33
    a[cm] = T[cm]/I[cm] = 5.33/.5 = 10.67
    and now
    A[t] = r*a[cm] = 2*10.67 = 21.33

    giving the same translational acceleration of the cylinder as before.

    So it still looks right to me.

    (Textbooks have been known to give incorrect answers occasionally.) :)
  12. May 15, 2003 #11
    I might be wrong because I just learned this stuff and it's still sinking in and I need lots of practice, but I don't think that you can use the Parallel Axis Theorem like that.

    If you choose the point p as your axis then the cylinder is still rotating about its cm as it rolls around p and you need to take that into account and add 1/2mr^2 to your equation, don't you?
  13. May 15, 2003 #12
    We do it just to aggravate the Canadians.

    Anyway, I suspect (based upon his "behaviour") that discoverer02 is not American.
  14. May 15, 2003 #13
    woops.. you did that didn't you.
  15. May 15, 2003 #14

    Remember, everything is relative to the frame of reference. The cylinder is rotating about its cm ONLY if you CHOOSE the cm as the axis of rotation. MY cylinder is rotating about point p (in my first set of equations), so I must use the moment of inertia about point p. Only when I shift my axis to the cm (as in my subsequent post) should I use the moment of inertia about the cm.
    Notice that both ways I get the same result for the translational motion.

    That's EXACTLY how the parallel axis theorem is used. (How else would you use it?)
  16. May 15, 2003 #15
    Anyway, I hope Tom or someone else who really knows this stuff will take a look at what I did & let us know if it is correct.
  17. May 15, 2003 #16
    Tension of 2.5 lb is consistent with acceleration of 2 ft/s^2 for the block. But it doesn't seem to make sense for the cylinder. If T=2.5, then the torque about the cm is 2.5*2=5 ft-lb.
    Then the angular acceleration a = T / I = 5/.5 = 10. Then the tangential acceleration A[t] = R*a = 2*10=20 ft/s^2
    F = m*A[t]
    F = (8/32)*20 = 5 lb
    saying that if the tangential (and translational) acceleration is 20 ft/s^2, then the net force on the cylinder must be 5 lb.

    But the cylinder weighs 8 lb, and the net force on the cylinder is
    the weight minus T. If T=2.5 lb, we get 8 - 2.5 = 5.5 lb, not 5 lb. So this seems to be inconsistent, and that's why I say T can't be 2.5 lb, and therefore A (for the block) can't be 2 ft/s^2. At least that's the way I see it.

    ...unless somehow the tangential acceleration of the circumference of the cylinder is NOT equal to the cylinder's vertical (translational) acceleration. But I don't see how that can be...
    Last edited: May 15, 2003
  18. May 15, 2003 #17
    I'm American too. I wasn't sure how to spell the word so I just sounded it out. :smile:

    Rotational motion and angular momentum are killing me.
  19. May 15, 2003 #18
    Behaviour is Brit for behavior. :smile:

    Anyway, my argument is summarized above (I just edited that post), so maybe you can check it with your teacher -- I won't see mine for a week & a half.
  20. May 15, 2003 #19
    The parallel axis theorem was presented to me as a way to determine the rotational inertia of an object about a point other than its center of mass, but in the examples I've seen so far, it was actually rotatating about that other point and not just imagined as the result coordinate axes placement.

    Like I said, it was just introduced to me this week and I need to review, digest and practice over the weekend.:smile:
  21. May 15, 2003 #20
    While you're digesting, digest this:
    there is no such thing as "actually" (in the sense that you are using it). Everything is relative to your choice of a frame of reference. I stand before you and declare that at any given instant in time, that cylinder is ACTUALLY rotating about point p. Prove me wrong!
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