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Torque or Force (ball rolling without slipping)?

  1. Jun 15, 2015 #1
    If there is a uniform ball rolling without slipping on an inclined plane, does gravity provide a torque, translational force, or both? I'm just really confused about forces vs. torques i guess?
     
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  3. Jun 15, 2015 #2

    blue_leaf77

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    If the ball is uniform, its center of gravity coincides with the rotation axis (the geometrical center), so the gravitational force exerts no torque on the ball.
     
  4. Jun 15, 2015 #3

    jbriggs444

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    Whether a force provides a torque depends on the axis of rotation you use. The force of gravity acts as if it were applied at an object's center of mass.

    If you use the ball's center of mass as the chosen axis of rotation then there is zero offset between the point of application and the axis of rotation. Zero moment arm means zero torque.

    If you use the point where the ball touches the inclined plane (the momentary center of rotation) as the chosen axis then there is a perpendicular offset between the point of application and the chosen axis of rotation. Non-zero force multiplied by non-zero perpendicular offset means non-zero torque.
     
  5. Jun 15, 2015 #4
    If you use the point where the ball touches the inclined plane (the momentary center of rotation) as the chosen axis then there is a perpendicular offset between the point of application and the chosen axis of rotation. Non-zero force multiplied by non-zero perpendicular offset means non-zero torque.[/QUOTE]

    I'm new and not sure how to properly quote the above but jbriggs444 posted that.

    Wow I never thought about a momentary center of rotation. I think I understand it now!

    This isn't a homework question I'm just trying to understand something, If i wanted to find the force of gravity with F=ma would i have to add the translational acceleration and the angular acceleration? Or do I ignore the translational acceleration? Or something else?
     
  6. Jun 15, 2015 #5

    jbriggs444

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    If you already know an object's mass then you can simply multiply by the acceleration of gravity (9.8 meters per second2 on the surface of the earth) to get the force that gravity exerts on it.

    The force of gravity can be added to all the other forces acting on an object (do you know how to draw a "free body diagram"?) to determine the net force on the object and therefore its translational acceleration.

    The torque from gravity can be added to all the other torques acting on an object to determine the net torque on the object (about the chosen reference axis). This will give you the rate of change of angular momentum. Angular momentum can be split into two parts:

    1. The rotation of an object around its center of mass.
    2. Movement of the center of mass relative to the chosen axis of rotation. Multiply the object's linear momentum by its mass and by the perpendicular offset from the chosen axis. [technically you are computing a vector cross product]

    If you already know the translational acceleration of the object (having done your free body diagram and added up the forces) and you know its offset from the chosen axis then you can calculate the rate of change of part 2. If you know all of the torques then you know how total angular momentum is changing. The difference is the rate at which angular momentum is accumulating in or being drained from the object's rotation. Divide by the moment of inertia and you have angular acceleration.

    It can be convenient to choose an axis of rotation that coincides with an object's center of mass. Then the second part of angular momentum is sure to be zero and all you have to worry about is the first part.
     
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