Question

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?

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