Unformatted text preview: 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 ) = 61.3 m/s b) The roof’s edge is at height H = 4.00 m. How far horizontally from the roof’s edge does the ball hit the level ground? The speed of the ball at the edge of the roof is v = r ω = 4.91 m/s At the edge, the ball is moving at an angle of 35º below the horizontal. x and y components of the intial velocity of the projectile motion is v 0x = v =cos(35º) = 4.02 m/s, v 0y = v sin(35º) = 2.81 m/s When the ball hits the ground, the vertical displacement of the ball is –H. The flight time is H = v 0y t – (1/2)gt 2 t = (v 0y + √ (v 0y 2 + 2gH))/g = 0.661 s During this time interval, the horizontal displacement of the ball is Δ x = v 0x t = 2.65 m...
View
Full Document
 Summer '08
 Field
 Conservation Of Energy, Work, Physical quantities, Energy Conservation Equation, Tomoyuki Nakayama, Discussion Section 0132H

Click to edit the document details