Rolling: A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal. How fast is it moving after it has rolled 2.2 m down the plane?

Question

Quizplus · Accepted Answer

To solve this problem, we will use conservation of energy. The potential energy at the top of the incline is equal to the sum of kinetic energy of the translational and rotational motion at the bottom of the incline. We can write this as:

$PE=KE_{trans}+KE_{rot}$

where $PE=mgh$ is the potential energy, $m$ is the mass, $g$ is the acceleration due to gravity, $h$ is the height of the incline, $KE_{trans}=\frac{1}{2}mv_{trans}^2$ is the translational kinetic energy and $KE_{rot}=\frac{1}{2}I\omega^2$ is the rotational kinetic energy. For a cylinder rolling without slipping, we have $v_{trans}=\omega R$, where $R$ is the radius of the cylinder. The moment of inertia for a solid cylinder about its center of mass is $I=\frac{1}{2}mR^2$.

Plugging in the given values, we have:

$mgh=\frac{1}{2}mv_{trans}^2+\frac{1}{2}I\omega^2$

Solving for $v_{trans}$ and simplifying, we get:

$v_{trans}=\sqrt{\frac{10}{7}gh}$

Substituting the given values, we get:

$v_{trans}=\sqrt{\frac{10}{7}	imes 9.8	imes 2.2	imes\sin 18^\circ}=3.0	ext{ m/s}$

Therefore, the answer is C.

Rolling: a Uniform Solid 5