Question

a solid cylinder of radius 11 cm and mass 20 kg starts from rest and rolls without slipping a distance L = 6.3 m down a roof that is inclined at angle θ = 32°. (a) What is the angular speed of the cylinder about its center as it leaves the roof? (b) The roof's edge is at height H = 5.3 m. How far horizontally from the roof's edge does the cylinder hit the level ground?

Answer 1

I think you're doing this with conservation of energy?

mgh = 1/2 mV^2 + 1/2mr^2 *omega^2
V= r*omega
\[2gh = (r \omega )^2 + r^2 \omega^2\]
\[gh = r^2 \omega^2\]

Answer 2

...where h is 5.3*sin 32

(b) is just a kinematics problem... find V from V=rw
Vy = -Vsin32
Vx = Vcos32