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July 31, 2014

July 31, 2014

Posted by **Joey** on Wednesday, January 5, 2011 at 9:33am.

(a) What is the angular speed of the cylinder about its center as it leaves the roof?

1 rad/s

(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?

2 m

- physics -
**drwls**, Wednesday, January 5, 2011 at 9:55amAre 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.

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