This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hinojosa (jlh3938) oldmidterm4 02 Turner (58185) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The oscillation of a massspring system x = x m cos( t + ) , where x m is a positive number. At time t = 0, the mass is at the equilibrium point x = 0 and it is moving with positive velocity v . What is the phase angle ? 1. = 3 2 correct 2. = 7 4 3. = 0 4. = 5. = 3 4 6. = 1 2 7. = 5 4 8. = 2 9. = 1 4 Explanation: Since the initial position is x = 0, we know that cos( ) = 0, so = 1 2 , 3 2 , 5 2 ,... . The velocity can be found as v = dx dt = x m sin( t + ) . At t = 0 we have v = x m sin( ) > 0, so the solution is = 3 2 . 002 10.0 points A solid cylinder of mass 4 kg and radius 35 cm is yoked to a spring as shown in the figure below. To be precise, the axle of the cylinder is attached to a horizontal spring of force constant 2658 N / m. The cylinder rolls back and forth on a horizontal base with out slipping. For simplicity, assume that the spring, the axle and the yoke which connects them have negligible masses compared to the cylinder itself. 2658 N / m 4 kg What is the angular frequency of the cylin der rolling back and forth around the equlib rium position? Correct answer: 21 . 0476 s 1 . Explanation: Let : M = 4 kg , R = 35 cm , and k = 2658 N / m . Consider the forces and the torques on the rolling cylinder: F s Mg N F f where F s = k x is the spring force and F f is the static friction force at the bottom of the cylinder. The directions of the forces correspond to x < 0; the spring is compressed and pushes to the right, so F s > 0. Since I cm = M R 2 2 , the torque equation around the cylinders axis gives = I cm = RF f M R 2 2 = RF f hinojosa (jlh3938) oldmidterm4 02 Turner (58185) 2 R = 2 F f M for the solid cylinder of uniform density. As long as the cylinder rolls without slipping, a x = R, so that M a x = M ( R ) = M 2 F f M = 2 F f . (2) The horizontal force equation gives us F net x = M a x = F s F f 2 F f = F x F f F f = 1 3 F x and M a x = F s F f = 2 3 F s , so a x = 2 k 3 M x. The equation of motion is harmonic, so a x d 2 x dt 2 = 2 x, which leads to simple harmonic motion with angular frequency = radicalbigg 2 k 3 M = radicalBigg 2 (2658 N / m) 3 (4 kg) = 21 . 0476 s 1 . 003 10.0 points A particle oscillates harmonically x = A sin( t + ) , with amplitude 13 m, angular frequency s 1 , and initial phase 3 radians. Every now and then, the particles kinetic energy and potential energy happen to be equal to each other ( K = U )....
View Full
Document
 Summer '08
 Staff
 Mass

Click to edit the document details