This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis is an unformatted preview. Sign up to view the full document.
View Full Documenthomework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A one-piece 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 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 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 = 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 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 .... View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentOld Midterm 3 - Solutions
Midterm 3 - Solutions
Homework 6 - Solutions
Physics Homework 1 - Solutions
Physics Homework 2 - Solutions
Homework 4 - Solutions
Homework 5 Solutions
Homework 6 Solutions
Homework 7 Solutions
Homework 8 Solutions
Homework 9 Solutions
Homework 10 Solutions
Copyright © 2015. Course Hero, Inc.
Course Hero is not sponsored or endorsed by any college or university.