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A uniform solid disk has a radius 1.60 m and a mass of 2.30 kg rolls...

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?

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Answer #1

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 <<<<<-----------answer

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