The center of mass goes down by the radius r
the change in potential energy is m g r
that is the kinetic energy
(1/2) I w^2
I is the moment of inertia about the pivot point which is the moment of inertia about the center + m r^2
the velocity v = w r
A uniform solid disk of radius 1.7 m and mass 81.3 kg is free to rotate on a frictionless pivot through a point on its rim.
If the disk is released from rest in the position shown by the solid circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle? The acceleration of gravity is 9.8 m/s^2.
The figure shows a solid circle, with a pivot point on the leftmost point of the rim. The dashed circle is basically rotated so that the pivot point is at the very top.
1 answer
1 answer