A thin solid disk of radius R = 0.5 m and mass M = 2.0 kg is rolling without slipping on a horizontal surface with a linear speed v = 5.0 m/s.The disk now rolls without slipping up an inclined plane that is at an angle of 60 degrees to the vertical.Calculate the maximum height that the disk rolls up the incline.
A)5.1 m
B)2.6 m
C)2.9 m
D)3.1 m
E)1.3 m

Question

Quizplus · Accepted Answer

The maximum height $h$ the disk rolls up the incline can be found using energy conservation. The initial kinetic energy (both translational and rotational) is converted into potential energy as the disk climbs the incline. The initial kinetic energy $K_{\text{initial}}$ is the sum of translational kinetic energy $\frac{1}{2}Mv^2$ and rotational kinetic energy $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the disk ($\frac{1}{2}MR^2$ for a solid disk) and $\omega$ is the angular velocity ($\frac{v}{R}$ for rolling without slipping). The potential energy at height $h$ is $Mgh$, where $g$ is the acceleration due to gravity (9.8 m/s$^2$). Setting the initial kinetic energy equal to the potential energy gives:$$\frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 = Mgh$$Simplifying and solving for $h$ gives:$$h = \frac{v^2}{2g}\left(1 + \frac{1}{2}\right) = \frac{5^2}{2 \times 9.8}\left(\frac{3}{2}\right) \approx 2.6 \text{ m}$$Therefore, the maximum height the disk rolls up the incline is approximately 2.6 meters.

A Thin Solid Disk of Radius R = 0