Posted by Joey on .
In the figure below, a solid cylinder of radius 12 cm and mass 11 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at angle è = 30°.
(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 = 5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?
Are 1 rad/s and 2 m your answers?
(a) I suggest you apply conservation of energy. The total kinetic energy is the sum of (1/2) M V^2 and (1/2) I w^2. where w = V/R.
w is the angular speed.
M g *L sin 30 = (1/2) M V^2 and (1/2) I w^2
= (1/2) M (Rw)^2 and (1/2)(1/2)MR^2) w^2
Solve for w. M cancels out
(b) Multiply the horizontal component of the velocity at the end of the roof by the time it takes to fall a distance H. Note that is leaves the roof with a downward velocity component.