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Unformatted text preview: kim (jkk547) HW07 Tsoi (58020) 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 dentists drill starts from rest. After 4 . 28 s of constant angular acceleration it turns at a rate of 17140 rev / min. Find the drills angular acceleration. Correct answer: 419 . 369 rad / s 2 . Explanation: Let : t = 4 . 28 s and f = 17140 rev / min . Since = 0, = f t = f t = 17140 rev / min 4 . 28 s 2 rev 1 min 60 s = 419 . 369 rad / s 2 . 002 (part 2 of 2) 10.0 points Throughout what angle does the drill rotate during this period? Correct answer: 3841 . 08 rad. Explanation: = t + 1 2 t 2 = 0 + 1 2 (419 . 369 rad / s 2 ) (4 . 28 s) 2 = 3841 . 08 rad . 003 (part 1 of 4) 10.0 points Consider the setup shown, where the inclined plane has a frictionless surface. The blocks have masses m 2 and m 1 . The pulley has mass m 3 , and is a uniform disc with radius R . Assume the pulley to be frictionless. m 1 T 1 R m 3 m 2 = T 2 a 30 Identify the equation of motion for m 1 . As sume the mass m 1 is more massive and is descending with acceleration a . The moment of inertia of the disk is 1 2 M R 2 . The accelera tion of gravity is 9 . 8 m / s 2 . 1. m 1 g sin  T 1 = m 1 a 2. T 1 m 1 g = m 1 a 3. T 1 + m 1 g sin = m 1 a 4. T 1 + m 1 g cos = m 1 a 5. T 1 + m 1 g = m 1 a 6. T 1 m 1 g cos = m 1 a 7. m 1 g cos  T 1 = m 1 a 8. T 1 m 1 g sin = m 1 a 9. T 1 + m 1 g = m 1 a 10. m 1 g T 1 = m 1 a correct Explanation: Using the freebody diagrams for the hang ing mass, summationdisplay F = T 1 m 1 g = m 1 a. 004 (part 2 of 4) 10.0 points Identify the equation of motion for m 2 . 1. T 2 m 2 g = m 2 a 2. m 2 g cos  T 2 = m 2 a kim (jkk547) HW07 Tsoi (58020) 2 3. m 2 g T 2 = m 2 a 4. T 2 + m 2 g = m 2 a 5. T 2 + m 2 g cos = m 2 a 6. m 2 g sin  T 2 = m 2 a 7. T 2 m 2 g sin = m 2 a correct 8. T 2 m 2 g cos = m 2 a 9. T 2 + m 2 g = m 2 a 10. T 2 + m 2 g sin = m 2 a Explanation: T 2 W bardbl N W W m 2 a W bardbl Applying Newtons law to m 2 , summationdisplay F = T 2 m 2 g sin = m 2 a. 005 (part 3 of 4) 10.0 points Identify the correct equation. 1. T 2 T 1 = 1 2 m 3 a 2. T 1 + T 2 = 1 2 m 3 a 3. T 2 T 1 = 7 10 m 3 a 4. T 1 = m 3 a 5. T 1 T 2 = 1 2 m 3 a correct 6. T 2 = m 3 a 7. T 2 T 1 = m 3 a 8. T 1 T 2 = 5 7 m 3 a 9. T 1 T 2 = m 3 a 10. T 1 + T 2 = m 3 a Explanation: a = R, so net = I ( T 1 T 2 ) R = parenleftbigg 1 2 m 3 R 2 parenrightbigg parenleftBig a R parenrightBig T 1 T 2 = 1 2 m 3 a. 006 (part 4 of 4) 10.0 points Find the acceleration a for the case where m 1 = 2 m 2 , the pulley is massless ( m 3 = 0) and = 30 ....
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This note was uploaded on 10/04/2009 for the course PHY 58020 taught by Professor Tsoi during the Fall '09 term at University of Texas at Austin.
 Fall '09
 TSOI
 Physics

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