A uniform cylinder I = mR²) of diameter 0.20 m and mass 12 kg rolls without slipping down a 37^o inclined plane. The gain in translational kinetic energy of the cylinder when it has rolled 5 m down the incline of the plane is approximately A) 24 J. B) 0.12 kJ. C) 0.24 kJ. D) 0.35 kJ. E) 0.59 kJ.

Question

Quizplus · Accepted Answer

The gravitational potential energy of the cylinder when it is located at the top of the incline plane with respect to its location at the bottom is given by:
$$\Delta U = mgh = mgsin	heta d = 12	imes9.81	imes sin37 	imes 5\approx 295.3 J$$
During its descent down the incline plane, the cylinder's mass moment of inertia and its rotational speed are constant. At any given point during its descent, the total kinetic energy of the cylinder is the sum of its rotational kinetic energy and its translational kinetic energy.
We can calculate the maximum rotational speed of the cylinder at the bottom of the incline plane as follows:
The gravitational potential energy at the top of the incline is equal to the translational kinetic energy at the bottom, plus the rotational kinetic energy at the bottom:
$$mgh = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$$
Since the cylinder is rolling without slipping, the velocity of its center of mass and the angular velocity of the cylinder are related as:
$$v=\omega R$$
where R is the radius of the cylinder. Hence the previous equation can be rewritten as:
$$mgh = \frac{1}{2}mv^2+\frac{1}{2}I\frac{v^2}{R^2}$$
Solving for v yields:
$$v=\sqrt{\frac{2mgh}{mR^2+I}}$$
Since the mass moment of inertia of the cylinder about its axis of symmetry is given by:
$$I=\frac{1}{2}mR^2$$
we can simplify the previous expression to:
$$v=\sqrt{\frac{4gh}{3}}\approx 2.77 m/s$$
The translational kinetic energy of the cylinder when it has rolled 5 m down the incline of the plane is given by:
$$K_t=\frac{1}{2}mv^2=\frac{1}{2}	imes 12	imes 2.77^2\approx 45.66 J$$
The gain in translational kinetic energy of the cylinder is then:
$$\Delta K_t = K_t - \frac{1}{2}mgsin	heta d\approx 45.66 - 295.3	imes sin37 	imes 5\approx 240.6 J\approx 0.24kJ$$
Therefore, the correct answer is (C).

A Uniform Cylinder I = MR2) of Diameter 0

A Uniform Cylinder I = MR²) of Diameter 0