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# HW9 - boland(kbb544 HW9 mackie(10611 This print-out should...

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boland (kbb544) – HW9 – mackie – (10611) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A can of soup has a mass 120 g, height 17 cm, and diameter 2 . 8 cm. It is placed at rest on the top of an incline that is 1 . 4 m long at 28 to the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . Calculate the moment of inertia of the can if it takes 0 . 88 s to reach the bottom of the incline. Correct answer: 6 . 40817 × 10 6 kg m 2 . Explanation: The speed of the center of mass of the can at the bottom of the incline is v f = 2¯ v = 2 L t . Thus the angular speed at the bottom of the incline is ω f = v f R . Using energy conservation, we have m g L sin θ = 1 2 m v 2 f + 1 2 I ω 2 f . Solving for I , I = 2 m ω 2 f bracketleftbigg g L sin θ - 1 2 v 2 f bracketrightbigg = 2 (120 g) (227 . 273 rad / s) 2 × bracketleftbigg (9 . 8 m / s 2 )(1 . 4 m) sin(28 ) - 1 2 (3 . 18182 m / s) 2 bracketrightbigg = 6 . 40817 × 10 6 kg m 2 . 002 10.0 points A solid steel sphere of density 7 . 5 g / cm 3 and mass 0 . 6 kg spin on an axis through its center with a period of 3 . 5 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 0 . 000307839 kg m 2 / s. Explanation: The definition of density is ρ M V = M 4 3 π R 3 , Therefore R = bracketleftbigg 3 M 4 π ρ bracketrightbigg 1 / 3 = bracketleftbigg 3 (0 . 6 kg) 4 π (7500 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0267301 m . Using ω = 2 π T = 2 π (3 . 5 s) = 1 . 7952 s 1 and I = 2 5 M R 2 = 2 5 (0 . 6 kg)(0 . 0267301 m) 2 = 0 . 000171479 kg m 2 , we have L I ω = 4 π M R 2 5 T = 4 π (0 . 6 kg)(0 . 0267301 m) 2 5 (3 . 5 s) = 0 . 000307839 kg m 2 / s . 003 10.0 points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone.

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boland (kbb544) – HW9 – mackie – (10611) 2 v 9 . 8 m / s 2 4 m 9 kg 19 Calculate the magnitude of the angular mo- mentum of the bob about a vertical axis through the supporting point. The acceler- ation of gravity is 9 . 8 m / s 2 . Correct answer: 24 . 5694 kg · m 2 / s. Explanation: Let : = 4 m , θ = 19 , g = 9 . 8 m / s 2 , and m = 9 kg . Consider the free body diagram. T m g θ Newton’s second law in the vertical and horizontal projections, respectively, gives T cos θ - m g = 0 T cos θ = m g and T sin θ - m ω 2 sin θ = 0 T = m ω 2 ℓ , where the radius of the orbit is R = sin θ .
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