A cylinder of mass 10 kg and radius 5 cm rolls from rest without slipping a dist

Question

A cylinder of mass 10 kg and radius 5 cm rolls from rest without slipping a distance L down a roof sloped at an angle of theta = 37 degree, as shown. a. Find the angular speed of the cylinder about its center just as it leaves the roof. Assume that d = 8m. b. Find the horizontal distance Delta x from the roof edge where the cylinder hits the ground, if the height H of the roof edge is 5.7 m. Assume that as it leaves the roof the angular speed of the cylinder about its center is 14 pi rad/s 44 rad/s.

Explanation / Answer

using energy conservation

initially there is only potential energy at the edge there is only kinetic energy

Ui = Uf

final cylinder have both rotational and translational kinetic energy

mgh = mv^2/2 + Iw^2/2

I = moment of inertia of cylinder = mr^2/2

we know v = w*r

mgh = 1/2 * mr^2*w^2 + 1/4 * mr^2*w^2

mgh = 3mr^2*w^2/4

w = (1/r)*sqrt(4gh/3)

h = 8*sin37

r = 0.05 m

w = 158.63 rad/s

part b )

given w = 44 rad/s

v = w*r = 2.2 m/s

vx = v*cos37 = 1.757 m/s

vy = v*sin37 = 1.324 m/s

x = vx*t

y = vy*t + 1/2*gt^2

y = 5.7 m

t = 0.95 s

x = vx*t

x = 1.67 m

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