Question

A solid, uniform disk of radius 0.250 m and mass 53.2 kg rolls down a ramp of length 4.70 m that makes an angle of 12.0° with the horizontal. The disk starts from rest from the top of the ramp.

(a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp. m/s

(b) Find the angular speed of the disk at the bottom of the ramp. rad/s

Answer 1

Using the conservation of energy of the system:

ΔPE = ΔKE ; the change in potential energy = the change in kinetic energy

taking the reference origin to be the top of the ramp:

mg(0) - mg(-h) = final KE ; since the initial KE = 0

here I have used h as the height of the ramp and this is given by Lsin(θ) ; where θ is the angle with the horizontal and L is the length of the ramp.

mgh = translational KE + rotational KE

mgh = (1/2)mv² + (1/2)*I*ω²

here v is the velocity of the centre of mass at the bottom, I is the moment of inertia of the cylinder therefore I = (1/2)mr² and ω is the angular velocity of the cylinder at the bottom.

Now assuming that the cylinder does not slip on the ramp
ω = v / r

So mgh = (1/2)mv² + (1/2)(1/2)mr²(v / r)²

mgh = (1/2)mv² + (1/4)mv²

gh = (3/4)v² ; rearranging for v

so v = 2*√(gh / 3)

as h = Lsin(θ)

v = 2*√(gLsin(θ) / 3)

Now using g = 9.81 m/s², L = 4.7 m and θ = 12°

v = 2*√(9.81*4.7*sin(12) / 3)

v = 3.573 m/s

(b) At the bottom of the ramp

ω = v / r

with r being radius of the cylinder given

ω = 3.573 / 0.25 = 14.29 rad/s