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Unformatted text preview: 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 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: K rot + K trans + U = 0 K rot + K trans + 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 = mg = mv 2 ( R- r ) , v 2 = g ( R- r ) . mg ( h + r ) = mg (2 R- r ) + 1 2 mv 2 + 1 2 I 2 , mg h + 2 mg r = 2 mg R + 1 2 mv 2 + 1 2 parenleftbigg 2 5 mr 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 mg , F x = + 10 7 2 R + r R- r mg 2. F y =- mg , F x = + 5 7 2 R + r R- r mg 3. F y =- mg , F x = + 10 7 2 R + r R- r mg correct 4. F y =- mg , F x =- 2 R + r R- r mg 5. F y =- 2 mg , F x = + 10 7 2 R + r R- r mg Explanation: When the sphere is initially at h = 3 R and finally at point P, conservation of energy gives mg (3 R + r ) = mg R + 1 2 mv 2 + 1 5 mv 2 v 2 = 10 7 (2 R + r ) g . yoon (jy4326) HW09 markert (58840) 2 Therefore, at point P, F y =- mg and F x = N = F c = + mv 2 R- r = + 10 7 2 R + r R- r mg . 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 R 3 , what is its angular momentum? Correct answer: 0 . 000708392 kg m 2 / s. Explanation: The definition of density is M V = M 4 3 R 3 , Therefore R = bracketleftbigg 3 M 4 bracketrightbigg 1 / 3 = bracketleftbigg 3 (0 . 9 kg) 4 (7850 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0301366 m ....
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This note was uploaded on 01/08/2010 for the course CH 53675 taught by Professor Bocknack during the Fall '09 term at University of Texas at Austin.

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