1
$\begingroup$

enter image description here

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

$\endgroup$
2
$\begingroup$

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.

enter image description here

$\endgroup$
  • $\begingroup$ 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 '14 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 '14 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? $\endgroup$ – Joshua Benabou Jan 14 '15 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 '15 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.