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Unformatted text preview: karna (pk4534) HW 12 coker (58245) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A particle is located at the vector position vectorr = (1 . 8 m) + (3 . 9 m) and the force acting on it is vector F = (3 . 5 N) + (3 . 4 N) . What is the magnitude of the torque about the origin? Correct answer: 7 . 53 N m. Explanation: Basic Concept: vector = vectorr vector F Solution: Since neither position of the par ticle, nor the force acting on the particle have the zcomponents, the torque acting on the particle has only zcomponent: vector = [ xF y y F x ] k = [(1 . 8 m) (3 . 4 N) (3 . 9 m) (3 . 5 N)] k = [ 7 . 53 N m] k . 002 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(0 . 8 m) , (6 m)]? Correct answer: 10 . 75 N m. Explanation: Reasoning similarly as we did in the pre vious section, but with the difference that relative to the point [(0 . 8 m) , (6 m)] the y component of the particle is now [ y (6 m)], we have vector = { [ x a ] F y [ y b ] F x } k = { [(1 . 8 m) (0 . 8 m)] [3 . 4 N] [(3 . 9 m) (6 m)] [3 . 5 N] } k = { 10 . 75 N m } k . 003 10.0 points A uniform horizontal rod of mass 1 . 5 kg and length 0 . 98 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = m 2 12 . . 98 m F 7 6 pivot 1 . 5 kg If a 5 . 2 N force at an angle of 76 to the hor izontal acts on the rod as shown, what is the magnitude of the resulting angular accelera tion about the pivot point? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 4 . 70298 rad / s 2 . Explanation: Let : = 0 . 98 m , m = 1 . 5 kg , = 76 , and F = 5 . 2 N . By the parallel axis theorem, the moment of inertia of a stick pivoted at the end is I = I cm + md 2 = 1 12 m 2 + m parenleftbigg 2 parenrightbigg 2 = 1 3 m 2 . The sum of the torques (counterclockwise ro tation defined as positive) is summationdisplay = F sin mg 2 = I F sin mg l 2 = parenleftbigg 1 3 m 2 parenrightbigg 6 F sin 3 mg = 2 m karna (pk4534) HW 12 coker (58245) 2 vector = 6 F sin 3 mg 2 m Since 6 F sin 3 mg = 6 (5 . 2 N) sin76 3 (1 . 5 kg) (9 . 8 m / s 2 ) = 13 . 8268 N , then bardbl vector bardbl = vextendsingle vextendsingle vextendsingle 13 . 8268 N vextendsingle vextendsingle vextendsingle 2 (1 . 5 kg) (0 . 98 m) = 4 . 70298 rad / s 2 ....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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