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Questions about rolling and direction of static friction?

  1. Nov 28, 2015 #1
    I have spent quite some time trying to get this, can someone please help me understand this?

    A plank is being pulled by a constant force F, it rests on top of two identitcal sylinders - there is rolling without slipping. What I dont understand is, why does F(p) - red arrow force, affect the acceleration of the center of mass of the cylinder. Isnt the only thing the red arrow F(p) does, to make it rotate - so it only affects its angular acceleration, and not the acceleration of center of mass of cylinders - while F(g) is the only force that affects the acceleration of center of mass.?

    And a second question:


    Disk rolls down an incline, here there is also rolling without slipping. But why in the world is the friction force pointing upwards. The disk is rolling clockwise ,shouldn't the friction force be pointing downwards - opposite of what's in the figure? Because on the first picture the cylinder is also rolling clockwise, and the friction force is acting along the same direction as the cylinder is moving - to the right.

    Attached Files:

  2. jcsd
  3. Nov 28, 2015 #2

    Doc Al

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    When a force acts on a body it always contributes to the net force, which determines the acceleration of the center of mass. Depending upon its point of application, it may also create a torque. It's not one or the other, but both.

    Since the disk rolls without slipping there must be a torque increasing the angular speed to match the increasing linear speed. Friction supplies that torque. Note that friction acts to reduce the acceleration of the center of mass as it rolls down the incline.
  4. Nov 28, 2015 #3
    So in the first picture F(g) will reduce the angular acceleration but increase acceleration of center of mass? And what if we looked at the wheel of a car - would the motor be providing torque, acceleration of center of mass, or both?

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  5. Nov 28, 2015 #4


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    It is informative to consider what would happen if you took away the friction from the ground. When you consider the angular and linear accelerations produced by the force applied tangentially at the top, you find that the instantaneous centre of rotation of the cylinders will be above the ground. So although the mass centres move forwards, the part of the cylinder touching the ground moves a bit backwards.
    Contrast this with what would happen if the horizontal force were applied at the centre of the cylinder, again with no ground friction. The whole cylinder would move forwards without rotating, so the direction of slip between cylinder and ground has reversed. Thus, allowing ground friction now, the direction of that frictional force has reversed.
    In between these cases, there is a point where an applied horizontal force would result in rolling contact yet no frictional force at ground. Finding this point is often set as an exercise.
    What it directly provides is torque. The wheels 'try' to rotate, but friction from the ground acts to oppose that, leading to a forward force on the wheels. The wheels would accelerate forwards, but are somewhat held back by the inertia of the car. The wheels are thus subjected to two opposing torques, one from the axle and one as a result of the opposition between the frictional force at the ground and the inertial force of the car. They don't quite balance, the difference leading to the rotational acceleration of the wheel.

    By the way, I don't like the term "rolling friction" used in the problem statement. It isn't friction. They mean rolling resistance, which comes from the imperfect elasticity of the wheels.
  6. Nov 29, 2015 #5
    Amazing, thank you.
    Last edited: Nov 29, 2015
  7. Nov 29, 2015 #6
    Just one thing; "The wheels are thus subjected to two opposing torques, one from the axle" . How can there be a torque from the axle if the axle is right at the center of the wheel - because haven't we chosen the center of the wheel to be the origin point, and torque is T = r x F - isnt our radious gonna be zero? Or am I just supposed to look away from the force supplied by the motor, and only think of the motor as providing torque?
  8. Nov 29, 2015 #7


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    As far as the axle is concerned, the motor supplies torque, not force, so there's no need to multiply it by a radius.
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