enter image description here

If we apply $F=ma$, there is no net force and as such there is no acceleration in the center of mass. But due to the friction acting, there is a torque which gives rise to an angular acceleration of the object ($τ=Iα$) and as a result you can find a tangential acceleration on the object. How come the same object has two different accelerations? Is there something I've missed or done wrong? Please explain how the object would move or not move.


Based on your free body diagram, yes, the center of mass would not experience any acceleration, but the object would rotate about the center due to the torque from force $f$.

However, based on what you have drawn, perhaps you have misunderstood how the forces are at work here. It looks like to me that you have a ball, (or disk, etc.) that is on an incline. Perhaps there is some string attached to its center to prevent the ball from rolling down the incline. Therefore, it looks like to me that $F$ is the component of the ball's weight along the incline, $f$ is a frictional force, and $F'$ is the tension in the string.

Now, if this is the right scenario, then in your case $f$ would actually be $0$. $F'$ would exactly balance out $F$, and there would be no acceleration and no rotation. Remember, static friction only opposes relative motion. If the string is holding the ball in place, then there is no impending relative motion, so $f=0$.

Of course I could be wrong. Maybe $f$ is static friction, but the incline has a conveyor belt on it so that $f\neq0$. Or maybe there is come other mechanism that is actually pulling on the ball on the incline.

| cite | improve this answer | |
  • $\begingroup$ What would happen if there is actually a mechanism which is pulling on the ball without assuming that a string is attached to it? $\endgroup$ – Himath Ipalawatta Jul 15 at 1:05
  • $\begingroup$ @HimathIpalawatta Well what is $F'$ supposed to be in that scenario? $\endgroup$ – BioPhysicist Jul 15 at 2:20
  • $\begingroup$ Assume something is pulling upwards which happens to equal F+f. $\endgroup$ – Himath Ipalawatta Jul 15 at 7:12
  • $\begingroup$ @HimathIpalawatta It doesn't matter if it's a string or something else. What is $f$ in this scenario? $\endgroup$ – BioPhysicist Jul 15 at 10:07
  • $\begingroup$ Limiting friction $\endgroup$ – Himath Ipalawatta Jul 17 at 4:05

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.