100points A pulley that has a moment of inertia 3 4 M...

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kondapaneni (sk43659) – Practice Midterm 02 – yao – (88200) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A pulley that has a moment of inertia 3 4 M R 2 and radius R rotates about an axis through its center. A string is wrapped around the disk and exerts a constant force 1 2 M g tangential to the disk. Find the tangential acceleration of a point on the rim of the disk. 1. a = 2 g 3 g A tire placed on a balancing machine in a ser- vice station starts from rest and turns through 4 . 72 rev in 1 . 63 s before reaching its final an- gular speed. Find its angular acceleration. 1. 25.3801 2. 16.3091 3. 21.4236 4. 218.864 5. 49.4801 6. 32.3701 7. 33.9194 8. 127.065 9. 137.711 10. 22.3242
2. a = 4 3. a = 2 g 5 4. a = g 5. a = 3 g 2 6. a = g 4 7. a = 2 g 3 correct 8. a = 4 g 3 9. a = g 3 10. a = g 2 Explanation: τ net = I α parenleftbigg 1 2 M g parenrightbigg R = parenleftbigg 3 4 M R 2 parenrightbigg α α = 2 g 3 R , so a = R α = 2 g 3 . 002 10.0points Explanation: Let : ω 0 = 0 rev / s , θ = 4 . 72 rev , and t = 1 . 63 s . There is a constant angular acceleration, so θ = ω 0 t + 1 2 α t 2 = 1 2 α t 2 α = 2 θ t 2 = 2 (4 . 72 rev) (1 . 63 s) 2 · 2 π rad rev = 22 . 3242 rad / s 2 . 003 10.0points Consider the elastic head-on collision between a sledge hammer with 7000 g mass and a golf ball with a 3 g mass. The initial velocity of the sledge hammer is 16 m / s , and the golf ball is initially at rest. 7000 g 3 g 16 m / s Estimate the approximate final speed v 2 of the golf ball.
kondapaneni (sk43659) – Practice Midterm 02 – yao – (88200) 2 1. v 2 80 m / s 2. v 2 5 . 09296 m / s 3. v 2 160 m / s 4. v 2 128 m / s 5. v 2 64 m / s 6. v 2 32 m / s correct 7. v 2 160 m / s 8. v 2 16 m / s 9. v 2 8 m / s 10. v 2 50 . 2655 m / s Explanation: v 1 = 16 m / s , v 2 = 0 m / s , m 1 = 7000 g , and m 2 = 3 g . Clearly the expression m 2 m 1 1. Therefore, we will approximate its square as zero. First, we conserve momentum: m 1 v 0 = m 1 v 1 + m 2 v 2 v 1 = v 0 m 2 m 1 v 2 Now, we place this formula into our equa- tion for conservation of energy, not bother- ing to calculate any terms proportional to parenleftbigg m 2 m 1 parenrightbigg 2 . 1 2 m 1 v 2 0 = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 m 1 v 2 0 = m 1 bracketleftbigg v 0 parenleftbigg m 2 m 1 parenrightbigg v 2 bracketrightbigg 2 + m 2 v 2 2 m 1 v 2 0 m 1 v 2 0 2 m 2 v 0 v 2 + m 2 v 2 2 0 ≈ − 2 m 2 v 0 v 2 + m 2 v 2 2 0 m 2 v 2 ( 2 v 0 + v 2 ) The v 2

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