A solid, uniform disk of radius 0.250 m and mass 54.1 kg rolls down a ramp of length 3.80 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.
(b) Find the angular speed of the disk at the bottom of the ramp.
Moment of inertia of solid disk , I = 1/2 M R^2
First , we will find the height of the ramp
h = 3.8*sin12 = 0.790064 m
By law of conservation of energy ,
Potential energy = Translational kineticenergy + 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 = sqrt( 4*9.8*0.790064/3)
b) )The angular speed of the disk at the bottom
of the ramp is
ω = v / R = 3.21302084/0.250 = 12.8520834 rad/s
let me know in a comment if there is any problem or doubts
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