A 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 4.27 rad/s at the bottom,what is the height of the inclined plane?
A)3.57 m
B)2.68 m
C)3.14 m
D)4.28 m

Question

Quizplus · Accepted Answer

The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. Using energy conservation: mgh = 0.5mv^2 + 0.5Iω^2. For a solid disk, I = 0.5mr^2. Solving for h gives h = (0.5v^2 + 0.25r^2ω^2)/g. With v = rω, substituting the given values results in h ≈ 3.57 m.

A Solid Disk of Radius 1