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Rotational kinematics - acceleration

  1. Oct 14, 2006 #1
    so there's a wheel with radius R and mass M. there's also a hub attached to the wheel's center with radius r and mass m. there's also a mass X suspended from a massless string that's wound around the hub. if the axle has negligible radius and mass and both wheel and the hub are solid with uniform density, how would you find the acceleration of the suspended mass after its released?

    i thought what it was askng for was the tangential acceleration, which i found to be equal to (F*r^2)/I.... (since tan acc is r*alpha where alpha is r*F/I because torque = I*alpha and also r*F)... so i tried doing:
    [r^2*X*g]/0.5[M*R+m*r]
    but that doesn't work and i'm not sure what im doing wrong... can someone point me in the right direction?
    thanks!
     
  2. jcsd
  3. Oct 14, 2006 #2

    Doc Al

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    Staff: Mentor

    Two problems:
    (1) The force F pulling on the hub does not equal the weight of the hanging mass. It does equal the tension in the string.
    (2) The rotational inertial of a disk is 0.5MR^2.

    Set up equations (Newton's 2nd law) for both wheel/hub and hanging mass and solve them together to get the acceleration.
     
  4. Oct 14, 2006 #3
    1- isn't the tension of the string the same thing as mg?

    also, what do you mean by "set up equations... to get the acceleration"? the 2nd law equation for the wheel/hub would be (m+M)alpha=F and for the hanging mass it would be F=Xg?...? i'm lost :(
     
  5. Oct 15, 2006 #4

    Doc Al

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    No. Think about it: if the tension in the string was equal to mg, then the net force on the mass would be zero. It would just sit there. (This is what would happen if you hung the mass from a string that was fixed to the roof, say.) But since that string moves, the tension will be less than mg.

    Here are the equations you need:
    (a) Torque = I alpha ==> Fr = I_total*alpha = I_total*a/r
    (b) Xg - F = Xa

    Note: We know that the acceleration is down, so I take down to be positive.
     
  6. Oct 15, 2006 #5
    I see, this is probably a dumb question but why did you go from Fr=I_total*alpha => F = I_total*a/r (that is, alpha to a?)

    and if you didn't mean to switch over... I know I'm supposed to be solving for acceleration, so how do i know what alpha is?

    lastly... I don't really understand the concept of "I"... the mass and radius of what object is supposed to be involved...? is it everything that's involved in the whole system? or just the wheel that's actually doing the turning? or just the hub that the string is directly attached to? or both?
     
  7. Oct 15, 2006 #6

    Doc Al

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    Why? Because we are trying to find "a", not alpha. Since the wheel/hub is connected to the hanging mass via the string, alpha can be related to "a" via: a = alpha*r (where r is the radius of the hub that the string wraps around).

    They are directly related. (See above.)

    The whole thing turns as one piece, so you must use the "I" for the entire wheel/hub object--which is just the sum of I_wheel and I_hub.
     
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