Kaplunovsky - hw 13

# Kaplunovsky - hw 13 - homework 13 – – Due 4:00 am 1...

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Unformatted text preview: homework 13 – – Due: Apr 25 2007, 4:00 am 1 Question 1, chap 11, sect 3. part 1 of 3 5 points A clock balance wheel has a period of os- cillation of 0 . 47 s. The wheel is constructed so that very nearly all of its 33 g of mass is concentrated around rim of radius 0 . 6 cm. What is the angular frequency of oscilla- tion? Correct answer: 13 . 3685 1 / s (tolerance ± 1 %). Explanation: Basic Concepts: For a physical pendu- lum, I α = τ = ⇒ I d 2 θ dt 2 =- κθ Solution: Given T , m , and r , f = 1 T = 1 . 47 s = 2 . 12766 cycles / s . The angular velocity ω is ω = 2 π f = 2 π (2 . 12766 cycles / s) = 13 . 3685 1 / s . Question 2, chap 11, sect 3. part 2 of 3 5 points What is the wheel’s moment of inertia? Correct answer: 1 . 188 × 10 − 6 kg m 2 (tolerance ± 1 %). Explanation: For a hoop I = mr 2 = (0 . 033 kg) (0 . 006 m) 2 = 1 . 188 × 10 − 6 kg m 2 . Question 3, chap 11, sect 3. part 3 of 3 8 points What is the torsion constant of the attached spring? Correct answer: 0 . 000212315 Nm (tolerance ± 1 %). Explanation: For torsional motion τ net = I α = I d 2 θ dt 2 =- κθ ω = radicalbigg κ I Therefore κ = I ω 2 = (1 . 188 × 10 − 6 kg m 2 ) (13 . 3685 1 / s) 2 = 0 . 000212315 Nm . Question 4, chap 15, sect 1. part 1 of 1 10 points A solid cylinder of mass M = 2 . 1 kg and radius R = 24 cm is yoked to a spring as shown in the figure below: 1343 N / m 24 cm 2 . 1 kg To be precise, the axle of the cylinder is at- tached to a horizontal spring of force constant k = 1343 N / m. The cylinder rolls back-and- forth on a horizontal base without slipping. For simplicity, assume that the spring, the axle and the yoke which connects them have negligible masses compared to the cylinder itself. What is the angular frequency of the cylin- der rolling back-and-forth around the equlib- rium position? Correct answer: 20 . 6482 rad / s (tolerance ± 1 %). Explanation: Consider the forces and the torques on the homework 13 – – Due: Apr 25 2007, 4:00 am 2 rolling cylinder: F s Mg N F f where F s =- kx is the spring force and F f is the static friction force at the bottom of the cylinder. The directions of all the forces cor- respond to x < 0: The spring is compressed and pushes to the right, thus F s > 0. The horizontal force equation gives us M a x = F net x = F s- F f (1) while the torque equation around the cylin- der’s axis gives I cm × α = R × F f (2) where I cm = MR 2 2 (3) for the solid cylinder of uniform density. Fi- nally, as long as the cylinder rolls without slipping, we must have a x = R × α (4) and therefore M a x = MR × α = MR × τ I cm = MR × R × F f 1 2 MR 2 = 2 × F f . (5) Substituting this equation into eq. (1) now gives us 2 F f = M a x = F s- F f (6) and hence F f = 1 3 F s , (7) M a x = 2 3 F s , (8) Finally, let us combine eq. (8) for the rolling cylinder with the spring equation F s =- kx : a x = 2 3 F s M =- 2 k 3 M × x. (9) This equation of motion is of harmonic type a x ≡ d 2 x dt 2 =- ω 2 ×...
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Kaplunovsky - hw 13 - homework 13 – – Due 4:00 am 1...

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