A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is 5.35 rad/s at the bottom, what is the height of the inclined plane?
A) 5.61 m
B) 4.21 m
C) 4.94 m
D) 6.73 m

Question

Quizplus · Accepted Answer

The correct answer is A because the disk rolls without slipping, so its linear velocity at the bottom of the inclined plane is v = rω = (1.60 m)(5.35 rad/s) = 8.56 m/s. The height of the inclined plane is then h = v^2/2g = (8.56 m/s)^2/(2 * 9.81 m/s^2) = 5.61 m.

A Uniform Solid Disk of Radius 1