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Conceptual question about torque

  1. Apr 24, 2015 #1
    1. The problem statement, all variables and given/known data
    This is a conceptual question about a prior assignment I had:

    A thin spherical shell rolls down an incline without slipping. If the linear acceleration of the center of mass of the shell is 0.23g, what is the angle the incline makes with the horizontal?
    2. Relevant equations
    m*g*r*sin(theta) = Icm Alpha

    3. The attempt at a solution
    This equation worked, but I was thinking about it after the fact, and I can't figure out why it worked. Torque = F*r, in this case the F is the force of gravity acting parallel to the incline, but that would only give me translational motion. The force causing torque comes from the friction between the plane and the wheel, which should be mg*u. So for this equation to work, there must be some other relationship between torque and translational motion. What I do know for a fact:
    1. Linear acceleration is equal to angular acceleration multiplied by the radius.
    2. The kinetic energies of rotation and translation are equal.
    So either there's another aspect of the relationship I'm missing, or the equation I used somehow accounts for the friction force.

    tl, dr: why does m*g*r*sin(theta) = Icm Alpha work when it should be m*g*u*r = Icm Alpha?
     
  2. jcsd
  3. Apr 24, 2015 #2

    SammyS

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    When calculating torque, you need to calculate it about some point.

    What is that point in your case?
     
  4. Apr 24, 2015 #3
    But I didn't have to calculate it, I set m*g*r*sin(theta) = 2/3 M r^2 a/r, and it simplified so m, g, and r all cancelled out. My question is just why is m*g*r*sin(theta) equal to torque at all? Doesn't the torque come from the friction between the plane and shell?
     
  5. Apr 24, 2015 #4

    SammyS

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    Well, that is calculating it symbolically. So, I assume that is with respect to the center of the sphere. Right?
     
  6. Apr 24, 2015 #5
    Yes. I don't think I'm asking this question well. If torque comes from the friction between the plane and the sphere, why is there no friction in the equation for torque?
     
  7. Apr 24, 2015 #6

    SammyS

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    There can be. If torque with respect to the center of the sphere is used to get angular acceleration, α, then it's the friction force producing the torque. But that's static friction so the actual amount applied is less than μsN, where N is the normal force, so μs doesn't appear in the result.

    The quantity mgRsin(θ), where R is the sphere's radius, is the torque about the point of contact of the sphere and incline. This can be used, to find α because the entire sphere has same angular acceleration, α .
     
  8. Apr 24, 2015 #7
    Oh, I think I get it. So I can view the torque as either a force of friction occurring at the point of contact (axis of rotation) that rotates the center OR a force that occurs at the center that rotates it around the point of contact (axis of rotation), right?
     
  9. Apr 24, 2015 #8

    haruspex

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    That's not how I'd put it.
    The friction force certainly occurs at the contact point, nowhere else. But torque and angular momentum depend on the reference axis you choose. If you take the centre of the ball/wheel as reference then there is an unbalanced torque from friction leading to angular acceleration. If you take your reference point as the point of contact then the friction force has no moment about that so does not appear in the equation. Instead, gravity has a moment, leading to the same angular acceleration.
    There's a frequently set problem where this is very handy. A ball is set sliding with given speed, no rotation, on a horizontal frictional surface. What speed does it have when it transits from sliding to rolling? By applying conservation of angular momentum about a reference point on the ground, friction is eliminated and the answer drops straight out.
     
  10. Apr 24, 2015 #9
    I get that gravity and friction balance so it rotates and translates, but I just don't understand why it is not included. m*g*sin(theta) is the force of gravity acting along the plane (translational force), I don't understand how just adding the radius makes it a rotational force. Let me use another example.

    A horizontal force of 10N is applied to a wheel of mass 10 kg and a radius 0.3 m. The wheel moves smoothly on the horizontal surface and the acceleration of the center of mass is 0.6 m/s^2. What is the rotational inertia of the rotational axis through its center of mass?
    To solve this equation (which seems strikingly similar to me), I can't use the applied force of 10N*r =Icm alpha. I need to use the Friction force* r = Icm*alpha.

    In the first problem, I can use the translational force, in the second problem, I can't. What's the difference here?
     
    Last edited: Apr 24, 2015
  11. Apr 24, 2015 #10

    SammyS

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    Without friction, gravity will not cause rotation for the sphere on the incline.
     
  12. Apr 24, 2015 #11
    I know, that's my question. Gravity will not cause rotation, only translation. So why is friction not included.

    Let me use another example.

    A horizontal force of 10N is applied to a wheel of mass 10 kg and a radius 0.3 m. The wheel moves smoothly on the horizontal surface and the acceleration of the center of mass is 0.6 m/s^2. What is the rotational inertia of the rotational axis through its center of mass?
    To solve this equation (which seems strikingly similar to me), I can't use the applied force of 10N*r =Icm alpha. I need to use the Friction force (-4)* r = Icm*alpha.

    In the first problem, I can use the translational force, in the second problem, I can't. What's the difference here?
     
  13. Apr 24, 2015 #12

    haruspex

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    There doesn't seem to be enough information. We're not told the height of the applied force, and since we are asked for the moment of inertia we can't assume a uniform disc etc. So there are three unknowns (force height, MoI, frictional force) and only two equations available.
     
  14. Apr 24, 2015 #13
    The force is applied to the center of the wheel in the diagram.
    10N moves an object of mass 10kg 0.6m/s^2, or 6N, so there is a force of friction moving in the opposite direction of -4N. 4N*r = Icm(.6/.3) so Icm = 0.6 kgm^2.

    So in the second question, I use the force of friction acting backwards as the force causing the rotation, in the first question, I use the translational force as the cause of rotation. Why?
     
  15. Apr 24, 2015 #14

    SammyS

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    If the wheel doesn't slip at the point of contact you can solve this.

    Moreover, you can solve this with translational force. The rotational inertia you get initially will be through an axis at the point of contact. Use the parallel axis theorem to find the rotational inertia for an axis through the center of mass. (Yes, we assume the wheel has radial symmetry.)

    Added in Edit:

    You do need to know where the force is applied, as harupex points out. I was assuming it was to be applied at the center of the wheel via an axle.
     
    Last edited: Apr 24, 2015
  16. Apr 24, 2015 #15
    I'm so lost I can't even ask the question clearly.
     
  17. Apr 24, 2015 #16

    SammyS

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    Generally these problem can be solved in several different ways.

    The ones you have suggested can be solved by taking friction into account rather directly or alternatively by taking it into account indirectly, for instance, by simply noting that there is no slipping at the point of contact.

    Furthermore, the methods discusses in this thread were all based on force/torque and acceleration. Even more fun can be had by finding solutions based on work and energy.
     
  18. Apr 24, 2015 #17

    haruspex

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    As SammyS says, given the height of the applied force you can solve it involving the frictional force or not, as you wish.
    If you take moments about the centre of the wheel, friction will come into the equation, so you will also need the equation for horizontal linear acceleration. But since you do not actually need to find the frictional force, it is more convenient to take moments about the point of contact.
    Again as SammyS says, you need to use the parallel axis theorem to relate the Icm (the unknown to be determined) to the Iperimeter.
     
  19. Apr 25, 2015 #18
    The effective mass ( in non slipping problem ) of thin shell sphere = m * 1.6667
    The acceleration = force / effective mass
    But
    The force = m * g * sine A
    And the acceleration = 0.23 * g
    So :
    0.23 * g = ( m * g * sine A ) / ( m * 1.6667 )
    Then :
    A = inverse sine ( 0.23 * 1.6667 )
     
  20. Apr 25, 2015 #19

    haruspex

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    Judging from the OP, pdg already got all that. You seem to have missed what he is asking.
     
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