A solid cylinder of mass m,radius r,and length L starts from rest and rolls without slipping down an inclined plane.Its vertical position decreases by a distance h.What is the final linear speed of its center of mass?

Question

Quizplus · Accepted Answer

The potential energy lost by the cylinder is converted into both translational and rotational kinetic energies. The work done by the gravitational force on the cylinder is given by mgh, and this must equal the increase in kinetic energy of the cylinder.

The rotational kinetic energy is given by (1/2)Iω^2, where I is the moment of inertia of the cylinder about its axis of rotation (I = (1/2)mr^2 for a solid cylinder) and ω is the angular velocity of the cylinder.

The translational kinetic energy is given by (1/2)mv^2, where v is the linear velocity of the center of mass of the cylinder.

Since the cylinder is rolling without slipping, the linear velocity of the center of mass is equal to the product of its angular velocity and its radius: v = ωr.

Setting the work done equal to the sum of the rotational and translational kinetic energies, we get:

mgh = (1/2)(1/2)mr^2ω^2 + (1/2)mv^2

Substituting v = ωr, we get:

mgh = (1/4)mr^2ω^2 + (1/2)mω^2r^2

Solving for ω, we get:

ω = sqrt(8gh/3r^2)

Substituting into v = ωr, we get:

v = r * sqrt(8gh/3r^2) = sqrt(8gh/3)

Therefore, the final linear speed of the center of mass of the cylinder is given by choice B.

A Solid Cylinder of Mass M,radius R,and Length L Starts