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solution_09pdf - yoon(jy4326 HW09 markert(58840 This...

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yoon (jy4326) – HW09 – markert – (58840) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A small, solid sphere of mass m and radius r rolls without slipping along the track con- sisting of slope and loop-the-loop with radius R at the end of the slope. It starts from rest near the top of the track at a height h , where h is large compared to r . P R m r θ h What is the minimum value of h (in terms of the radius of the loop R ) such that the sphere completes the loop? The moment of inertia for a solid sphere is 2 5 m r 2 . 1. h min = 2 . 5 ( R - r ) 2. h min = 1 . 8 ( R - r ) 3. h min = 2 . 8 ( R - r ) 4. h min = 2 . 7 ( R - r ) correct 5. h min = 3 . 7 ( R - r ) Explanation: Δ K rot + Δ K trans + Δ U = 0 Δ K rot + Δ K trans + Δ U = 0 Initially the center of mass of the sphere is a distance h + r above the bottom of the loop and as the mass reaches the top of the loop, this distance above the reference is 2 R - r . Conservation of energy gives h = h min when the speed of the sphere at the top of the loop satisfies the condition summationdisplay F = m g = m v 2 ( R - r ) , v 2 = g ( R - r ) . m g ( h + r ) = m g (2 R - r ) + 1 2 m v 2 + 1 2 I ω 2 , m g h + 2 m g r = 2 m g R + 1 2 m v 2 + 1 2 parenleftbigg 2 5 m r 2 parenrightbigg parenleftBig v r parenrightBig 2 g h + 2 g r = 2 g R + 7 10 v 2 = 2 g R + 7 10 g ( R - r ) = 2 ( R - r ) + 0 . 7 ( R - r ) = 2 . 7 ( R - r ) . 002 (part 2 of 2) 10.0 points What are the force components on the sphere at the point P, which has coordinates ( - R, 0) if we take the center of the loop as origin, and if h = 3 R ? 1. F y = - 1 2 m g , F x = + 10 7 2 R + r R - r m g 2. F y = - m g , F x = + 5 7 2 R + r R - r m g 3. F y = - m g , F x = + 10 7 2 R + r R - r m g correct 4. F y = - m g , F x = - 2 R + r R - r m g 5. F y = - 2 m g , F x = + 10 7 2 R + r R - r m g Explanation: When the sphere is initially at h = 3 R and finally at point P, conservation of energy gives m g (3 R + r ) = m g R + 1 2 m v 2 + 1 5 m v 2 v 2 = 10 7 (2 R + r ) g .
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yoon (jy4326) – HW09 – markert – (58840) 2 Therefore, at point P, F y = - m g and F x = N = F c = + m v 2 R - r = + 10 7 2 R + r R - r m g . 003 10.0 points A solid steel sphere of density 7 . 85 g / cm 3 and mass 0 . 9 kg spin on an axis through its center with a period of 2 . 9 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 0 . 000708392 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 . 9 kg) 4 π (7850 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0301366 m . Using ω = 2 π T = 2 π (2 . 9 s) = 2 . 16662 s 1 and I = 2 5 M R 2 = 2 5 (0 . 9 kg)(0 . 0301366 m) 2 = 0 . 000326958 kg m 2 , we have L I ω = 4 π M R 2 5 T = 4 π (0 . 9 kg)(0 . 0301366 m) 2 5 (2 . 9 s) = 0 . 000708392 kg m 2 / s . 004 (part 1 of 2) 10.0 points A student sits on a rotating stool holding two 4 . 4 kg masses. When his arms are extended horizontally, the masses are 0 . 89 m from the axis of rotation, and he rotates with an angu- lar velocity of 1 . 5 rad / sec. The student then
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