Suppose, we have a wheel, $I = kmR^2$ ($k$ is any coefficient for a shape), of radius R and the torque is applied by the axle at radius r because it is spinning. We shouldn't consider that applied torque to be a linear force because it really isn't. There will be a force at the bottom that is caused by the torque and it will cause the friction that (if rolling without slipping) should be equal to the force at the bottom. Suppose $F_{app}$ is the force component of a torque $ \frac {T_{app}} r$ applied by the axle and $F_r$ is force of friction. The friction force should be: $\frac{F_{app}r}R$ or $\frac {T_{app}}R$. The friction force applies no torque but acts as the only force applied. So the equations are:

$T = I\alpha \implies T_{app} = F_{app} r = \frac {kmR^2a}R$

$\implies a = \frac{F_{app}r} {kmR} = \frac {F_r} m$

Is that not true?

Now the real problem is how do we calculate the acceleration if the the force caused by the torque applied is too large for the friction and so kinetic friction will be there and slipping will occur. Help would be very appreciated.

  • $\begingroup$ Usually when we say "A torque is applied", we mean that it is created such that it's net force is $0$ (e.g. think of turning a wheel with your two hands on opposite sides of the wheel). Is this what you are assuming here for the applied torque? $\endgroup$ – BioPhysicist Apr 29 at 16:08
  • $\begingroup$ yes. It is what we could call a couple. I think. $\endgroup$ – GameOver Apr 29 at 16:16
  • $\begingroup$ So then what is $F_{app}$? And how is friction not applying a torque? $\endgroup$ – BioPhysicist Apr 29 at 16:21
  • $\begingroup$ AHH... I am really confused now. the case is simple, the axle is spinning the wheel, how do we calculate the acceleration if there is frcition. $\endgroup$ – GameOver Apr 30 at 3:58
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    $\begingroup$ You can apply a similar analysis as in my answer to your previous question :) $\endgroup$ – BioPhysicist Apr 30 at 4:06