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Unformatted text preview: homework 10 – FRENNEA, KYLE – Due: Apr 6 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A onepiece cylinder is shaped as in the figure below, with a core section protrud ing from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, of radius 1 . 29 m, exerts a force F 1 to the right on the cylinder. A rope wrapped around the core, of radius 0 . 413 m, exerts a force F 2 downward on the cylinder. x y O R F 1 R 2 1 F 2 Let F 1 = 4 . 59 N and F 2 = 8 . 59 N What is the net torque acting on the cylinder about the rotation axis (which is the z axis in the figure)? Correct answer: − 2 . 37343 N m (tolerance ± 1 %). Explanation: The torque due to F 1 is − R 1 F 1 and is neg ative because it tends to produce a clockwise rotation. The torque due to F 2 is + R 2 F 2 and is positive since it tends to produce a counter clockwise rotation. Therefore, the net torque about the rotation axis is τ net = R 2 F 2 − R 1 F 1 = − 2 . 37343 N m . Question 2 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 = [1 + k ] mv 2. K = bracketleftbigg 1 + 2 k 2 bracketrightbigg mv 2 3. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 4. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 5. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 correct 6. K = bracketleftbigg 1 + 2 k 2 bracketrightbigg mv 7. K = [1 + k ] mv 2 8. K = [1 + 2 k ] mv 2 9. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 10. K = [1 + 2 k ] mv Explanation: Basic Concepts: Rotational kinetic en ergy is E rot = 1 2 I ω 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 homework 10 – FRENNEA, KYLE – Due: Apr 6 2007, 4:00 am 2 also v . Then the angular velocity of the wheel is ω = v R . So the total kinetic energy is K = 1 2 2 mv 2 + 1 2 I ω 2 = 1 2 2 mv 2 + 1 2 k m ( Rω ) 2 = 1 2 2 mv 2 + 1 2 k mv 2 = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 . Question 3 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 8 7 g h 3. v = radicalbigg 8 11 g h 4. v = radicalbig 2 g h 5. v = radicalbigg 8 9 g h 6. v = radicalbigg 16 11 g h 7. v = radicalbigg 8 5 g h correct 8. v = radicalbig g h 9. v = radicalbigg 4 3 g h 10. v = radicalbigg 2 3 g h Explanation: The mechanical energy is conserved. So 2 mg h = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 = 5 4 mv 2 ....
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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