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Calculating Torque of sphere resting on an inclined plane

  1. Feb 23, 2005 #1
    If I have a sphere with mass m and radius r, resting on an inclined plane with angle [tex]\theta[/tex]. What is the torque on the sphere due to gravity pull?

    I know that the force that pulls the sphere down the plane is [tex]mg *\sin \theta[/tex] and the (static) frictional force is [tex]-\mu_s * mg \cos \theta[/tex].

    If we can assume the sphere does not slip, how does one calculate the torque that the frictional force exerts on the sphere?

    Im also interested in how to calculcate the total acceleration of the sphere.

    /CandyShot
     
    Last edited: Feb 23, 2005
  2. jcsd
  3. Feb 25, 2005 #2
    If you draw a neatly labeled freebody diagram, you will see that the force component of weight along the incline (downhill) is [itex]mg\sin\theta[/itex]. Now, this force acts through the mass center so it has a zero torque contribution about the CM. On the other hand (assuming of course that the incline doesn't translate) the friction force equal to f does not act through the mass center so it has a torque about the mass center which tends to increase the rotational velocity of the body (draw arrows to convince yourself that this is true).

    For a formal solution, consider [itex](i',j',k')[/itex] unit vectors along the incline, perpendicular to it ([itex]k' = i' X j'[/itex]). Resolve the forces and the vector [itex]\vec{r}[/itex] from the CM to the point of application of friction force and compute the torque due to this force by using [itex]\tau = \vec{r}X\vec{F}[/itex].

    For the total acceleration you need to write 2 dynamic equations: one for translation of the center of mass (as you would write for a block sliding down an incline) and one for rotation.

    Now it is important to make a distinction between rolling and slipping. The formula for friction force you have used can be applied only when slipping is given to occur. If rolling without slipping occurs then the the velocity of the point of contact of the body is zero. In this case you cannot use [itex]f = \mu mg\cos\theta[/itex] and must not substitute any value of f. f will enter the dynamic equation for rotation and also the translational equation. Eliminate f to get the relationship between the mass, theta and acceleration.
     
    Last edited: Feb 25, 2005
  4. Feb 25, 2005 #3
    Okey, thanks for that.

    /Candyshot
     
  5. Feb 25, 2005 #4
    I discovered a major mistake in my post and have edited it. Please read the post all over again. Sorry for the inconvenience.
     
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