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Torque/Intertia: Balancing a Tire

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    [tex]\Sigma[/tex][tex]\tau[/tex]= r x F = I*[tex]\alpha[/tex]

    Angular Momentum
    L= r x p = r x mv = r x mr[tex]\omega[/tex] = mr2 * [tex]\omega[/tex]2

    This formatting looks weird. It should be r^2 *(omega)^2

    and I*(alpha for the next part)


    3. The attempt at a solution

    Right now I'm working on part a

    I was thinking the I[tex]\alpha[/tex] of each wheel would give me the center of mass
    Torque = I \alpha + I\alpha
    but I don't have an acceleration value, and that wont give me where the center of mass. I think I need to find the center of mass first, and then figure out how adding the mass to one wheel will change the axis of rotation.
  2. jcsd
  3. Mar 10, 2009 #2

    Doc Al

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    Staff: Mentor

    Yes, figure out the center of mass first. Find the combined center of mass of the "perfect tire" of mass M and the extra mass m. (Where's the center of mass of the "perfect tire" by itself?)
  4. Mar 10, 2009 #3
    Well for the center of mass of the tire, I don't really have a radius for it?

    The mass m is 1 ft away from the center of the tire, but how far away is mass M from the center of the tire? Do I use 1 ft for it a well?

    But then I would get (M+m)/(M+m) which is just 1...


    is the center of mass equation, with just a general m and r in there
    Last edited: Mar 10, 2009
  5. Mar 10, 2009 #4
    Well, I could say that the perfect tire's R for the center of mass equation is 0, and that would just give me 0 as the center of mass. Then the center of mass for the two objects would be:

    and that would give me my new center of mass, which would be the distance between the center of mass and the axis of rotation

    I'm not sure how to continue with part B though.
  6. Mar 11, 2009 #5

    Doc Al

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    Staff: Mentor


    The center of mass moves in a circle, so what's its acceleration?
  7. Mar 11, 2009 #6
    Torque= R cross F= I times alpha

    Alpha is then a/R. But I need a velocity in there somehow to solve part C. 50mph isn't an angular velocity, nor does a/R contain velocity in it.

    How do I account for the change in the center of mass? Its going to involve my answer to A somehow, which I got to be a really small number. Is it just 1 + this number?

    edit: I need a way to relate a/R to linear velocity
    Last edited: Mar 11, 2009
  8. Mar 11, 2009 #7

    Doc Al

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    Staff: Mentor

    Parts A, B, and C have nothing to do with torque.

    To solve B, think of all the mass being located at the center of mass. Thus the mass is rotating in a circle of known radius (from part A). What force must be exerted to make it go in a circle? (Hint: Centripetal.)
  9. Mar 11, 2009 #8
    Looking a wikipedia

    1. (2piR)/V=T

    2. (2piV)/a=T

    3. a=(v^2)/R

    and for it in radians per second
    4. omega=2pi/T

    So do I just want equation 3, substitute it into F=ma,

    and get (M+m)(v^2/R)=answer for C?

    It says "frequency of the vibration" though, which doesn't sound like a force answer

    Well now I see omega=2pi/T=V/R <--- is that all I need? I feel like I need to take this and substitute it into something else


    I'm thinkin that I want this.

    I'm not sure how to use R though. Is it just simply my answer for part A, or is it 1 +/- my answer in part A? (Which I got to be 3/803)
    Last edited by a moderator: Apr 24, 2017
  10. Mar 11, 2009 #9

    Doc Al

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    Staff: Mentor

    Yes, where v is the tangential speed of the center of mass and R is the distance from the axis to the center of mass.

    Part C asks for both force and frequency. What's the frequency of rotation of the tire?
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