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Unformatted text preview: homework 24 HETTIGER, CHRISTOF Due: Mar 27 2008, 4:00 am 1 Question 1, chap 13, sect 1. part 1 of 2 10 points Consider a circular wheel with a mass m , and a radius R . The moment of inertia about the center of the wheel is I = k mR 2 , where k is a constant in the range between 0 . 5 k 1 . . A rope wraps around the wheel. A weight of mass 2 m is attached to the end of this rope. m R 2 m At some moment, the weight is falling with a speed v . The total kinetic energy K of the system at this moment is given by 1. K = bracketleftbigg 1 + 2 k 2 bracketrightbigg mv 2. K = [1 + k ] mv 3. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 4. K = [1 + 2 k ] mv 5. K = bracketleftbigg 1 + 2 k 2 bracketrightbigg mv 2 6. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 7. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 correct 8. K = [1 + 2 k ] mv 2 9. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 10. K = [1 + k ] mv 2 Explanation: Basic Concepts: Rotational kinetic en ergy is E rot = 1 2 I 2 K tot = K wheel + K mass At the moment the weight is descending with a speed v , the tangential speed of the wheel is also v . Then the angular velocity of the wheel is = v R . So the total kinetic energy is K = 1 2 2 mv 2 + 1 2 I 2 = 1 2 2 mv 2 + 1 2 k m ( R ) 2 = 1 2 2 mv 2 + 1 2 k mv 2 = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 . Question 2, chap 13, sect 1. part 2 of 2 10 points Assume: k = 1 2 . If the system is released from rest, find the speed v at the moment when the weight has descended a vertical distance h . 1. v = radicalbigg 4 5 g h 2. v = radicalbigg 2 3 g h 3. v = radicalbigg 8 7 g h 4. v = radicalbigg 4 3 g h 5. v = radicalbigg 16 11 g h 6. v = radicalbig 2 g h 7. v = radicalbigg 8 5 g h correct 8. v = radicalbigg 8 11 g h 9. v = radicalbigg 8 9 g h homework 24 HETTIGER, CHRISTOF Due: Mar 27 2008, 4:00 am...
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 Spring '10
 Turner
 Physics, Inertia, Mass, Work

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