1
$\begingroup$

enter image description here

I read that if the left block moves up by $dx$ then the center of mass of the rotating object moves down by a vertical length of $\frac{1}{2}dx\sin(\phi)$ but I thought it would move down by $dx\sin(\phi)$ as doesn’t the disk rotate down a length of $dx$ around its circumference?

$\endgroup$
0
$\begingroup$

This is a common confusion of beginners. Since the disk is doing pure rotation, the contact point of the disk and the slope can be treated as static. Then, instantly, you can think of the disk rotating around this contact point instead of its own center. Thus, the vertical length should be $\frac{1}{2}\mathrm{d}x\sin{\phi}$.

$\endgroup$

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.