The moment of inertia of a certain cylinder,whose mass is not distributed uniformly,is 0.6mR² about its geometric axis.The translational speed of the center of mass after it starts from rest and rolls 14 m down a 30º incline is approximately A)9.3 m/s B)86 m/s C)13 m/s D)3.1 m/s E)41 m/s

Question

Quizplus · Accepted Answer

The translational speed of the center of mass can be found using energy conservation principles. The total mechanical energy at the top of the incline is equal to the potential energy, given by $mgh$, where $h$ is the height of the incline. As the cylinder rolls down the incline without slipping, this energy is converted into both 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 and $\omega$ is the angular velocity. Since the cylinder rolls without slipping, $v = r\omega$, where $r$ is the radius of the cylinder.Given the incline angle $\theta = 30^\circ$ and the distance $d = 14 m$, we can find the height $h = d \sin(\theta)$. The moment of inertia is given as $0.6mR^2$, where $R$ is the radius of the cylinder. The energy conservation equation can be set up as follows:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(0.6mR^2)\left(\frac{v}{R}\right)^2$$Solving this equation for $v$ gives us the translational speed of the center of mass. The exact numerical solution depends on the values of $m$, $g$, and $R$, but given the options and the context, the correct answer is found through calculation based on the principles described, leading to an approximate speed of $9.3 m/s$.

The Moment of Inertia of a Certain Cylinder,whose Mass Is