A uniform solid disk has a radius 1.60 m and a mass of 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity is 4.09 rad/s at the bottom, what is the height of the inclined plane?
ROLLING(disk)
initial potential energy ui = m*g*h
initial kinetic energy ki = 0
total initial energy Ei = Ui + ki
Ei = m*g*h
at the bottom of the incline
final rotational kinetic energy KEr =
0.5*I*w^2
I = (1/2)*M*R^2
w = v/R
KEr = 0.5(1/2)*m*R^2*w^2 = 1/4*m*R^2*w^2
final translational kinetic energy KEt = 0.5*m*v^2 = (1/2)*m*R^2*w^2
final potential energy uf = 0
final energy Ef = (1/4)*m*R^2*w^2 + (1/2)*m*R^2*w^2
Ef = (3/4)*m*R^2*w^2
from conservation of energy
Ef = Ei
(3/4)*m*R^2*w^2 = m*g*h
(3/4)*1.6^2*4.09^2 = 9.8*h
h = 3.28 m
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