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If a sphere rolls on an inclined plane, why does the component of gravity parallel to the plane not exert a torque on the sphere? Shouldn't infinitesimal mass particles experience a force of (dm)gsin(theta)? Because infinitesimal mass particles experience the force at different lengths, shouldn't they experience torques and, consequently, give rise to a net torque?

I ask this question, because the following link (starting at subheader "Perspective II")

treats friction as the only torque-contributing force applied to an object on an incline.

http://physics.bu.edu/~duffy/sc527_notes06/race.html

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That's actually a really interesting question. Stating it in a slighly different way:

If there is no friction, would the sphere on the inclined plane experience a torque?

And the answer is - no it would not. The reason for this is that the external force on the sphere (the normal force of the plane) acts through the center of the sphere. In other words - there is no "arm" and thus no torque.

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  • $\begingroup$ To be a little more precise, the force of gravity creates no torque, if you treat the center of mass as the pivot about which we calculate torque. If we treat the point of contact with the ramp as the pivot, then the friction and normal force create no torque, and the force of gravity does. $\endgroup$
    – G. Paily
    Nov 8, 2014 at 23:04
  • $\begingroup$ @G.Paily - correct. I answered a slightly different question, which is "would a sphere start to roll down an incline if no friction was present" and demonstrate that the answer is "no". But your clarification is useful. $\endgroup$
    – Floris
    Nov 8, 2014 at 23:07
  • $\begingroup$ If there is torque due to gravity, then why doesn't rotation occur about the pivot point at the point of contact? $\endgroup$
    – math_lover
    Jan 14, 2015 at 0:04
  • $\begingroup$ @JoshuaBenabou - gravity alone doesn't produce torque since it acts through the center of gravity. You need a force that doesn't act through the c.o.g. - and the normal force doesn't qualify either. That leaves only friction. $\endgroup$
    – Floris
    Jan 14, 2015 at 0:15

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