Unformatted text preview: edge. E i = E f ⇒ mg(Lsin θ ) = (1/2)mv 2 + (1/2)I ω 2 mgLsin θ = (1/2)m(r 2 ω 2 ) + (1/2)(2/5)mr 2 ω 2 = (7/10)mr 2 ω 2 ω = √ (10gLsin θ /7r 2 ) = 147 rad/s b) The roof's edge is at height H = 5.00 m. How far horizontally from the roof's edge does the cylinder hit the level ground? After the sphere leaves the roof, it undergoes a projectile motion. The initial speed of the projectile motion is the speed of the sphere at the edge of the roof: v = r ω = 7.35 m/s At the edge, the sphere is moving at an angle of 40º below the horizontal. x and y components of the initial velocity of the projectile motion are v 0x = v =cos(-40º) = 5.63 m/s, v 0y = v sin(-40º) = -4.72 m/s When the sphere hits the ground, the vertical displacement of the sphere is –H. The flight time is -H = v 0y t – (1/2)gt 2 t = (-v 0y + √ (v 0y 2 + 2gH))/g = 0.637 s During this time interval, the horizontal displacement of the sphere is Δ x = v 0x t = 3.59 m...
View Full Document
This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08