A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of $25 ^ { \circ }$ with the horizontal. What is the forward speed of the disk after it has rolled 3.0 m, measured along the plane?
A) 4.1 m/s
B) 6.3 m/s
C) 3.5 m/s
D) 5.7 m/s
E) 2.0 m/s

Question

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of $25 ^ { \circ }$ with the horizontal. What is the forward speed of the disk after it has rolled  3.0 m, measured along the plane?
A)  4.1 m/s
B)  6.3 m/s
C)  3.5 m/s
D)  5.7 m/s
E)  2.0 m/s

Quizplus · Accepted Answer

The forward speed of the disk can be found using energy conservation principles. The potential energy lost by the disk as it rolls down the incline is converted into both translational and rotational kinetic energy. The potential energy lost is given by $mgh$, where $h$ is the vertical height descended, which can be found from the distance rolled down the plane, $s = 3.0 \, \text{m}$, and the angle of the incline, $\theta = 25^\circ$, using $h = s \sin \theta$.The total kinetic energy at the bottom 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 and $\omega$ is the angular velocity. For a solid disk, $I = \frac{1}{2}mr^2$, and since the disk rolls without slipping, $v = r\omega$.Setting the potential energy equal to the total kinetic energy and solving for $v$ gives us the forward speed. The calculation simplifies to $v = \sqrt{2gh}$. Plugging in the values, we find that $v$ is approximately 4.1 m/s, making choice A correct.

A Uniform Solid Disk Is Released from Rest and Rolls 25∘25 ^ { \circ }25∘

A Uniform Solid Disk Is Released from Rest and Rolls $25 ^ { \circ }$