0
$\begingroup$

Say a disc is kept on an incline. If the force of friction acting on it equals the component of gravitational force along the incline, the disc must not have linear acceleration, but then the force of friction exerts a torque about the body's center of mass.

If it has some angular acceleration but zero linear acceleration , the disc would just perform pure rotational motion about it's center of mass.

I just can't visualise a disc rotating about it's center of mass without moving down, where am I going wrong?

Improve this question
$\endgroup$
2
  • $\begingroup$ How is it kept on an inclined plane? Something else must be keeping it there. What causes angular acceleration is that the sum of torques is not null, and it moves downwards bc the gravitational component is not countered by another force. $\endgroup$
    – rmhleo
    Commented Apr 18, 2021 at 17:07
  • $\begingroup$ You're letting it loose.. $\endgroup$
    – Ali Sahad
    Commented Apr 18, 2021 at 17:36

1 Answer 1

Reset to default
0
$\begingroup$

If the block is free to move, the static friction will adjust itself so that a = Rα and it will be less than the component of gravity. If some external force prevents motion, the friction force will be zero. If the disk were on a fixed axle, your condition could be met by moving the the surface of the incline up along its length.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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