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qz8sol_0132Hf10

# qz8sol_0132Hf10 - mg(Lsin θ =(1/2)mv 2(1/2)I ω 2 mgLsin...

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TA: Tomoyuki Nakayama Monday, October 25, 2010 PHY 2048: Physic 1, Discussion Section 0132H Quiz 8 (Homework Set #9) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A solid ball of radius 8.00 cm and mass 5.00 kg starts from rest and rolls without slipping a distance L = 3.00 m down a roof that is inclined at angle θ = 35.0º. a) What is the angular speed of the ball about its center as it leaves the roof? While the ball is rolling down the roof, the mechanical energy is conserved. We apply the energy conservation equation to the initial position and the edge of the roof and solve the equation for the angular velocity at the edge. . E i = E f
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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...
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