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Rotational Wheel

  1. Apr 14, 2008 #1
    In the figure below, a constant horizontal force Fapp of magnitude 15 N is applied to a wheel of mass 14 kg and radius 0.40 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.43 m/s2.

    (a) In unit-vector notation, what is the frictional force on the wheel?

    (b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

    Ok so the wheel is rolling smoothly which means that static friction is involved. I came up that Frictional Force = uN = u(14 * 9.8) = 137.2 u

    So the only forces acting upon this object is the applied force, 15N and Friction which is 137.2 u.

    Now I'm lost...I don't know where to go from here...any help please?
  2. jcsd
  3. Apr 15, 2008 #2


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    Actually, a wheel rolling smoothly just means there is no slippage occurring. Think about what has to be true if the wheel is not to slip while moving translationally. Also this relates to kinetic, not static friction, as the wheel is in motion.
  4. Apr 15, 2008 #3


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    Hi Nanuven,

    In this case I don't believe you can use that formula for the frictional force. For the static frictional force

    u_s N

    gives the maximum static frictional force, which occurs when the surfaces are just about to slip. That does not appear to be happening here.

    I think to find the frictional force, perhaps use a force diagram and Newton's law, since the acceleration is given.


    Hi Nabeshin,

    When the wheel is rolling without slipping the kinetic frictional force is not involved. The difference between kinetic and static friction is whether or not the two surfaces are in relative motion at the contact point. But the bottom of a wheel that is in contact with the ground has zero velocity if it is rolling smoothly, so there is no relative motion at that point.
  5. Apr 15, 2008 #4


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    Hi Nanuven! :smile:

    You don't need the coefficient of friction at all.

    Just follow alphysicist's advice, and use Newton's second law: total horizontal force = mass times horizontal acceleration! :smile:
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