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Torque and acceleration Please help!

  1. May 12, 2005 #1
    For a disc with mass m, radius R, and moment of inertia about center of mass of (mR^2)/3, and applied external force is F.

    A wheel is being pulled by a force F directly upwards, which causes it to roll without slipping (due to static friction), on a horizontal surface. The upward force F is applied at the most right tip or edge of the disc.

    Find the acceleration, a, and the static frictional force, f.

    I'm not sure how to set up the torque or force equations etc....

    So, if F is directly upward, and on the edge, is it like F*cos90=0, so
    and second equation,
    F*(2R)=(moment of inertia about point of contact of disc and surface) * (angular speed)
    moment of inertia about point of contact = (moment of inertia about center of mass, given) + m*R^2
    angular speed=a/R (because no slip?)

    so in this case, solving the F*2R= --- etc. equation, we get
    and using 0-f=ma,

    Does this sound right at all or what should I do? Any suggestion is much appreciated!!!
  2. jcsd
  3. May 12, 2005 #2

    Doc Al

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    Staff: Mentor

    Welcome to PF!

    Is the disk nonuniform? For a uniform disk [itex]I_{cm} = 1/2 m R^2[/itex].
    The only horizontal force on the disk is the frictional force, so f = ma.
    Both friction (f) and the upward force (F) will exert a torque about the center of mass. Set the net torque equal to [itex]I \alpha[/itex]. (Yes, [itex]a = \alpha R[/itex], since it doesn't slip.)

    Redo your solution with my suggestions.
  4. May 12, 2005 #3
    Thank you!


    Equation 1:
    (where I is the the moment of inertia going through center of mass, i.e. 1/2mR^2) <--- so we just choose the torque equation to go through the center of mass, is that the easiest case? cos i remember in class we talked about picking the point of contact and use the moment of inertia with respect to that point, like using parallel axis theorem or something... that doesn't apply here does it? on what cases then do we do that?)

    Equation 2:
    -f=ma=m(r*alpha) <--- do we need the negative sign here? (cos that will affect our result a bit wouldn't it?)

    So just solve the two equations two unknowns to get the answer.

    (If I used it withOUT the negative sign, I get a=2F/3m, and f=2F/3, does this sound right? the acceleration and friction should be in the opposite directions right, or no..? do we just state it out or what do I do?
    cos otherwise, WIth the negative sign, I would have gotten a=-2F/m, and f=-2F, which can't be right... right?)

  5. May 12, 2005 #4

    Doc Al

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    Staff: Mentor

    You can certainly choose the contact point if you like. That makes the equation even easier: [itex]FR = I_{contact} \alpha = 3/2 m R^2 \alpha[/itex].

    The negative sign just depends on your choice of coordinates. I chose positive to be to the left, so both a and [itex]\alpha[/itex] are positive. (Otherwise I can't use [itex]a = I \alpha[/itex].)


    These answers look good to me. The acceleration must be in the same direction as the frictional force, since the friction causes the acceleration! (The moral of this story: always draw a careful diagram and identify all the forces and accelerations.)

    No good. I'll bet you got your signs mixed up somewhere.
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