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A block on an inclined plane does not rotate?

  1. Feb 17, 2010 #1

    s39

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    A block of mass m is on a frictionless inclined plane with no initial speed.
    The net force which equals to the component of the block's weight along the inclined plane, acted through the block's centre of mass, so there is no torque acted on the block, so the block does not rotate and just accelerate downwards.
    But if the inclined plane has friction, and the magnitude of the friction is smaller than the magnitude of the component of the weight of the block along the plane, we know that the friction does not act through the centre of mass of the block, so there is a torque acted on the block due to the friction, but how come we cannot see any rotation of the block, the motion of the block is in fact accelerating downwards?
     
  2. jcsd
  3. Feb 17, 2010 #2
    If the coefficient of friction is the same at all points of contact between the block and the plane, there is be no net torque. But if the coefficient varies then there may be a torque and rotation. Try putting a drop of oil under one end of the block; it will probably rotate as well as slide down the ramp.
     
  4. Feb 17, 2010 #3

    s39

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    Sorry...can you further explain the above sentence, I don't understand why there is no net torque when the coefficient is the same at all points of contact.
     
  5. Feb 18, 2010 #4

    Doc Al

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    Staff: Mentor

    Don't forget that the incline also exerts a normal force on the block. That normal force is not uniformly distributed along the surface, but can adjust itself (up to a point) so as to prevent the block from passing through the incline. While friction exerts a torque in one direction, the normal force exerts a counter torque.
     
  6. Feb 18, 2010 #5
    I was thinking about rotation around an axis normal to the plane. Thanks to Doc Al, I understand my error. It's an interesting question and merits some experimentation. For example, are there any cases (e.g. block shapes) that would cause a sliding block to tumble if the coefficient of sliding friction varies down the plane? I must confess I don't know.
     
  7. Feb 19, 2010 #6

    s39

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    So is my free body diagram referring to the case correct?
    [shown in attached document]

    Even though the normal force is not evenly exerted on the block, the total normal force exerted on the block is still mgcos(A). And the net normal force should not act through the centre of mass of the block in order to balance the torque due to friction.
     

    Attached Files:

    Last edited: Feb 19, 2010
  8. Feb 19, 2010 #7

    Doc Al

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    Looks good to me.

    Exactly.
     
  9. Feb 19, 2010 #8

    s39

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    Thanks a lot! :)
     
  10. Feb 19, 2010 #9
    Whether rotation is possible or not depends on the angle of the inclined plane.
    If you observe from the frame of the forward point of contact of block, then the torque will be provided by mgsinQ and normal reaction.....in this case, friction does not matter...whereas if you observe from the frame of center of mass, friction provides the torque whose value again depends on the inclination angle.
     
  11. Feb 19, 2010 #10

    Doc Al

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    True, but here we know the block doesn't rotate. The issue is simply to understand how the torques balance.
    Careful about taking torques about an accelerating point that isn't the center of mass. You must add the torque due to an inertial force acting at the center of mass. You are correct that the friction exerts no torque about that point, but friction certainly does matter, regardless of reference frame. And you must include the torque due to the entire weight, not just the component parallel to the incline. The torques due to the weight, normal force, and inertial force add to zero.
     
  12. Feb 20, 2010 #11

    s39

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    What is inertial force? And in this case, how does inertial force act on the block?
     
  13. Feb 21, 2010 #12

    Doc Al

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    Newton's 2nd law only applies in an inertial frame. In order to use it in a non-inertial frame, such as a frame attached to the accelerating block, we must add so-called 'fictitious' or inertial forces. This is just an artifact of viewing things from an accelerating frame of reference.

    In this case, since the block accelerates down the incline, an inertial force equal to 'ma' going up the incline must be added. That 'force' acts at the center of mass. If we use the center of mass as our reference point for calculating torque, we can ignore the inertial force since it acts at that point and thus its torque is zero. But for any other reference point, we need to consider the torque due to the inertial force in addition to that of the other (real) forces acting.
     
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