Question

a 10 kg disk (I+1/2R^2) starts to roll w/o slipping down a 5m tall incline. what is its liner velocity at the bottom of the incline?

answer = 8.1 m/s

pls show work and steps

Answer 1

inital energy Ei = m*g*h

final energy energy at the bottom Ef = KEt + KEr

KEt = translational kinetic energy = (1/2)*m*v^2

KEr = rotatioanl kietic energy = (1/2)*I*w^2

I = momet of inertia= (1/2)*m*r^2

w = angular speed = v/r

therefore

Ef = (1/2)*m*v^2 + (1/2)*(1/2)*m*r^2*(v/r)^2

Ef = (1/2)*m*v^2 + (1/4)*m*v^2

Ef = (3/4)*m*v^2

from energy conservation

total energy is constant

Ef = Ei

(3/4)*m*v^2 = m*g*h

v = sqrt(4gh/3)

v = sqrt(4*9.81*5/3)

v = 8.1 m/s