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Finding the friction force for a rotating body

  1. Nov 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A solid cylinder is dragged over a surface by a constant force F that has an angle θ with horizontal. It rolls without friction around its symmetrical axis. The mass is M and radius is R and moment of inertia is [itex]I=\frac{1}{2}MR^2[/itex]
    The cylinder rolls on the surface without gliding.

    Use Newton's 2nd law for the center of mass and the rotation equation to find the friction force.


    2. Relevant equations
    Center of mass:
    [itex]\sum F_z=Ma_{CM}[/itex]

    Rotation equation:
    [itex]\sum \tau_z=I_{CM}\alpha_z[/itex]


    3. The attempt at a solution
    FBD:
    fbYXHsg.gif

    Is this FBD correct?

    I'm not sure what forces I have here because both the F force and normal force have a line of action through the center so they have no torque. What did I draw wrong?
     
  2. jcsd
  3. Nov 28, 2013 #2

    cepheid

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    Your problem statement says it rolls *without* friction, but then asks you to calculate the friction. Can you clarify this please?
     
  4. Nov 28, 2013 #3
    cepheid, I think he means no kinetic friction, because the cylinder rolls without gliding, so only the static friction.
     
  5. Nov 28, 2013 #4
    I believe what they mean is that it is a "pure rolling" where there isn't any equation of the form [itex]f=\mu n[/itex]
    They want the friction force on the cylinder. So there is still a friction force "f" that has to be considered in Newton's second law.

    I believe this is so because the normal force's line of action goes through the center of mass?
     
  6. Nov 28, 2013 #5

    cepheid

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    Sure, but the line of action of the friction force does not pass through the centre.

    I think the approach for this problem is to find the acceleration of the centre of mass using the force balance equation. For rolling without slipping, the linear and angular accelerations are related to each other in a simple way, so that will help you with the torque balance equation.

    Right, thanks.
     
  7. Nov 28, 2013 #6
    By the force balance equation you mean this right?
    [itex]\sum F_z=Ma_{CM}[/itex]

    So I have:
    [itex]a_{CM}=\frac{\sum F_z}{M}[/itex]

    My mass is known but I don't know what forces that act on the cylinder. Would it be correct to write that:
    [itex]\sum F_z=F-f[/itex]

    ??
     
    Last edited: Nov 28, 2013
  8. Nov 28, 2013 #7

    cepheid

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    Yes, that is what I meant by the force balance equation. I'm confused by why you have a subscript z on your force. If z is the long axis of the cylinder (i.e. it comes out of the page), then this subscript makes sense for the torque equation, because you're summing the torques that are in the z-direction. However, for the force, none of the forces are in the z direction, they all lie in the xy-plane (the page). You have two summations, one for x, and one for y. If you were trying to sum the horizontal (x) forces, it should not be F - f, since F is not entirely horizontal. You have to resolve it into components.
     
  9. Nov 28, 2013 #8
    Yeah, I understand what you mean. To be fair, I just copied exactly what my professor wrote during my lecture. Maybe he made a mistake. :)

    And you're absolutely right about the component. For some reason I completely forgot about that.

    [itex]a_{CM}=\frac{F\cdot cos(\theta)-f}{M}[/itex]

    I know that the relation between angular and linear acceleration is [itex]a=R\cdot \alpha[/itex]

    We know the radius, so we can express the linear acceleration in terms of the angular acceleration:

    [itex]R\cdot \alpha=\frac{F\cdot cos(\theta)-f}{M}[/itex]

    This can be inserted into our rotation equation:

    [itex]\sum \tau_z=I_{CM}\frac{\frac{F\cdot cos(\theta)-f}{M}}{R}[/itex]

    We know that the moment of inertia for the solid cylinder:

    [itex]\sum \tau_z=\frac{1}{2}MR^2\frac{\frac{F\cdot cos(\theta)-f}{M}}{R}[/itex]

    Now there's only the torque forces left.
    My guess is that only the friction force has a torque?

    So that leaves:

    [itex]f\cdot R=\frac{1}{2}MR^2\frac{\frac{F\cdot cos(\theta)-f}{M}}{R}[/itex]

    And then of course the equation can be simplified and then everything is known except for the friction force.

    Does this sound correct?
     
  10. Nov 28, 2013 #9

    cepheid

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    It looks okay to me, algebraically. I believe you are correct that only the friction force causes a torque in this situation.
     
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