Question

A solid disk (radius R=2.5 cm , mass M =0.35 kg) rolls without slipping down an 30 degree-incline. If the incline is 4.2 m long and the disk starts from rest, what is the linear velocity of its center of mass at the bottom of the incline (in m/s)?

Answer 1

The initial energy of the solid disk, Ei = M * g * h
Where M is the mass of the disk and h is the height of the incline.
h = L * sin$\theta$
Where L is the length of the incline and $\theta$ is the angle of incline.
h = 4.2 * sin(30)
= 2.1 m

Final energy of the disk = Translational kinetic energy + Rotational kinetic energy.
Ef = 1/2 * M * V^2 + 1/2 * I * $\omega$ ^2
Where V is final velocity of the center of mass, I is the moment of inertia, and $\omega$ is the angular velocity.
For a disk, I = 1/2 * M * R^2
Where R is the radius of the disk
Since the disk was rolling without slipping, $\omega$ = V/R
Ef = 1/2 * M * V^2 + 1/2 * [1/2 * M * R^2] * (V/R)^2
= 1/2 * M * V^2 + 1/4 * M * V^2
= 0.75 * M * V^2

Using the conservation of energy, Ei = Ef
M * g * h = 0.75 * M * V^2
V^2 = (g * h) / 0.75
= (9.8 * 2.1) / 0.75
= 27.44
V = SQRT[27.44]
= 5.2 m/s