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Physics HW7 - kim(jkk547 – HW07 – Tsoi –(58020 This...

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kim (jkk547) – HW07 – Tsoi – (58020) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A dentist’s drill starts from rest. After 4 . 28 s of constant angular acceleration it turns at a rate of 17140 rev / min. Find the drill’s angular acceleration. Correct answer: 419 . 369 rad / s 2 . Explanation: Let : t = 4 . 28 s and ω f = 17140 rev / min . Since ω 0 = 0, α = ω f - ω 0 t = ω f t = 17140 rev / min 4 . 28 s · 2 π 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: θ = ω 0 t + 1 2 α 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 μ = 0 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 θ - T 1 = m 1 a 2. T 1 - m 1 g = m 1 a 3. T 1 + m 1 g sin θ = m 1 a 4. T 1 + m 1 g cos θ = m 1 a 5. T 1 + m 1 g = m 1 a 6. T 1 - m 1 g cos θ = m 1 a 7. m 1 g cos θ - T 1 = m 1 a 8. T 1 - m 1 g sin θ = 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 free-body 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 θ - T 2 = m 2 a
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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 θ = m 2 a 6. m 2 g sin θ - T 2 = m 2 a 7. T 2 - m 2 g sin θ = m 2 a correct 8. T 2 - m 2 g cos θ = m 2 a 9. T 2 + m 2 g = - m 2 a 10. T 2 + m 2 g sin θ = m 2 a Explanation: T 2 W bardbl N W W m 2 a W bardbl θ Applying Newton’s law to m 2 , summationdisplay F = T 2 - m 2 g sin θ = 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 α , so τ net = I α ( 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 θ = 30 . 1. a = 7 8 g 2. a = 3 4 g 3. a = 5 8 g 4. a = 1 2 g correct 5. a = 1 8 g 6. a = 1 4 g 7. None of these 8. a = 3 8 g Explanation: Let : θ = 30 . T 1 - T 2 = 1 2 m 3 a = 0 T 1 = m 1 g - m 2 a = 2 m 2 g - 2 m 2 a
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kim (jkk547) – HW07 – Tsoi – (58020) 3 T 2 = m 2 a + m 2 g sin θ = m 2 a + 1 2 m 2 g T 1 - T 2 = 3 2 m 2 g - 3 m 2 a 0 = 3 m 2 g - 6 m 2 a a = 1 2 g .
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