A solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at the angle theta = 30degree.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
(b) The roofs edge is at height H = 5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?

Explanation:

Acceleration of cylinder

a = g sin 30 / 1+ k² / r² where k is radius of gyration and r is radius of cylinder.

For cylinder k²  = (1 / 2)  r²

acceleration

= gsin30 / 1.5

= g / 3

= 3.27

v² = u² + 2as

= 2 x 3.27 x 6

v = 6.26 m /s

v = angular velocity x radius

6.26 = angular velocity x .10

angular velocity  = 62.6 rad / s

b ) vertical component of velocity

= 6.26 sin 30

= 3.13 m /s

h = ut + 1/2 g t²

5 = 3.13 t + .5 t²

.5 t²+ 3.13 t- 5 = 0

t = 1.32 s

horizontal distance covered

= 6.26 cos 30 x 1.32

= 7.15  m

it is a negative acceleration