Question

A solid, uniform disk of radius 0.250 m and mass 60.6 kg rolls down a ramp of length 4.80 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

Given

Mass of the disk, M = 60.6 kg

Length of the ramp , L = 4.80 m

Radius of the disk , R = 0.250 m

Angle of inclination, θ = 12 degrees

Moment of inertia of solid disk , I = 1/2 M R^2

a) First , we will find the height of the ramp

     sin (12) = h / 4.80 m

                h = 0.99798 m

By law of conservation of energy ,

    Potential energy = Translational kinetic energy + Rotational kinetic energy

     Mgh = 1/2 Mv^2 + 1/2 I ω2

     Mgh = 1/2 Mv^2 + 1/2 (1/2 M R^2)ω2            [ since v = Rω]

    Mgh =    1/2 Mv^2 + 1/4 Mv^2

    Mgh = 3/4 M v^2

            v = √(4 gh / 3)

            v = √(4 *9.8 m/s^2 *0.99798 m / 3)

            v = 3.61 m/s is the speed of the disk at the bottom of the ramp

b)The angular speed of the disk at the bottom of the ramp is

         ω = v / R = 3.61 m/s / 0.250 m

         ω = 14.44 rad /s