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PHY 303K - Old Midterm #3

# PHY 303K - Old Midterm #3 - oldmidterm 03 FIERRO JEFFREY...

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oldmidterm 03 – FIERRO, JEFFREY – Due: Apr 1 2008, 11:00 pm 1 Question 1, chap 12, sect 2. part 1 of 1 10 points The tub of a washer goes into its spin- dry cycle, starting from rest and reaching an angular speed of 3 . 3 rev / s in 4 . 6 s . At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 6 . 9 s . Through how many revolutions does the tub turn? Assume Constant angular acceleration while it is starting and stopping. Correct answer: 18 . 975 rev (tolerance ± 1 %). Explanation: We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period. While speeding up, θ 1 = ω t = 0 + ω 2 t 1 = 1 2 ω t 1 = 1 2 (3 . 3 rev / s) (4 . 6 s) = 7 . 59 rev . While slowing down, θ 2 = ω t = ω + 0 2 t 2 = 1 2 ω t 2 = 1 2 (3 . 3 rev / s) (6 . 9 s) = 11 . 385 rev . So θ = θ 1 + θ 2 = (7 . 59 rev) + (11 . 385 rev) = 18 . 975 rev . Question 2, chap 13, sect 2. part 1 of 1 10 points Consider a wheel of mass m , outer radius R , unknown but radially symmetric mass dis- tribution, and moment of inertia 3 7 mR 2 . The wheel is given a spin around its axis, and then it’s carefully lowered on a horizontal surface. (The wheel does not bounce.) At the moment of contact, the wheel has angular velocity ω 0 but zero linear velocity, v 0 = 0, so it slips against the surface. The kinetic friction force between the sur- face and the wheel slows down the rotation of the wheel and at the same time gives it a hor- izontal acceleration. Eventually, the wheel’s linear motion catches up with its rotation, and the wheel begins to roll without slipping. I = 3 7 m R 2 R , radius m ω 0 μ Once the wheel rolls without slipping, what is its angular speed? 1. ω rot = 1 3 ω 0 2. ω rot = 2 5 ω 0 3. ω rot = 5 12 ω 0 4. ω rot = 3 11 ω 0 5. ω rot = 2 9 ω 0 6. ω rot = 3 8 ω 0 7. ω rot = 7 16 ω 0 8. ω rot = 3 10 ω 0 correct 9. ω rot = 3 7 ω 0

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oldmidterm 03 – FIERRO, JEFFREY – Due: Apr 1 2008, 11:00 pm 2 10. ω rot = 6 13 ω 0 Explanation: Once the wheel lays on the surface, there are 3 forces acting on it: the weight mg , the normal force N , and the kinetic friction force f . The weight and the normal force acts along the vertical line through the wheel’s center (which is where the center of mass is), so they create no torque around the center of mass. On the other hand, the friction force acts perpendicularly to the wheel’s radius at the distance R from the center, so it creates a torque τ = R × f. (1) Consequently, the wheel’s rotation decelerates at the rate α = τ I = Rf I , (2) ω ( t ) = ω 0 αt. (3) At the same time, the friction force gives the wheel a horizontal linear acceleration a = f m , (4) so at time t the wheel’s linear speed is v = at = f m × t. (5) The wheel’s linear motion catches up with its rotations when v ( t ) = R × ω ( t ) . (6) According to eqs. (3) and (5), this happens when at = 0 Rαt. (7) Solving this equation, we find t = 0 a + , (8) at which time the wheel rotates at angular velocity ω ( t ) = ω 0 α × 0 a + = ω 0 × a a + αR .
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PHY 303K - Old Midterm #3 - oldmidterm 03 FIERRO JEFFREY...

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