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Unformatted text preview: homework 16 FISHER, CHRISTIAN Due: Mar 24 2008, 11:00 pm 1 Question 1, chap 12, sect 2. part 1 of 1 10 points A phonograph record has an initial angular speed of 33 rev/min. The record slows to 10 rev/min in 1.7 s. What is the records average angular accel eration during this time interval? Correct answer: 1 . 4168 rad / s 2 (tolerance 1 %). Explanation: Basic Concept: avg = 2 1 t Given: 1 = 33 rev / min 2 = 10 rev / min t = 1 . 7 s Solution: avg = 2 1 t = 10 rev / min 33 rev / min 1 . 7 s 2 rad 1 rev 1 min 60 s = 1 . 4168 rad / s 2 Question 2, chap 11, sect 3. part 1 of 1 0 points A uniform disk of radius 2 . 5 m and mass 2 . 2 kg is suspended from a pivot 0 . 275 m above its center of mass. The acceleration of gravity is 9 . 8 m / s 2 . axis Find the angular frequency for small os cillations. Correct answer: 0 . 917618 rad / s (tolerance 1 %). Explanation: Basic Concepts The physical pendulum: = I = mg d sin = d 2 dt 2 so that the angular frequency for small oscil lations (sin ) is = radicalbigg mg d I . Parallel axis theorem I = I + ma 2 Solution: We need the moment of inertia of the disk about the pivot point, which we call P. The moment of inertia of a uniform disk about its center is I disk = 1 2 mR 2 , but here the disk is rotating about P, a dis tance d from the center of mass. The parallel axis theorem lets us move the axis of rotation a distance d : I P = 1 2 mR 2 + md 2 = m parenleftbigg R 2 2 + d 2 parenrightbigg . Then using the formula for the small angle oscillation frequency of a physical pendulum (see Basic Concepts above), we obtain = radicalBigg mg d I P = radicaltp radicalvertex radicalvertex radicalvertex radicalbt mg d m parenleftbigg R 2 2 + d 2 parenrightbigg or = radicaltp radicalvertex radicalvertex radicalvertex radicalbt g d R 2 2 + d 2 = radicaltp radicalvertex radicalvertex radicalvertex radicalbt (9 . 8 m / s 2 ) (0 . 275 m) (2 . 5 m) 2 2 + (0 . 275 m) 2 = 0 . 917618 rad / s . homework 16 FISHER, CHRISTIAN Due: Mar 24 2008, 11:00 pm 2 Question 3, chap 12, sect 2. part 1 of 3 10 points What is the tangential acceleration of a bug on the rim of a 65 . 7 rpm record of diameter 15 . 4 in . if the record moves from rest to its final angular speed in 3 . 35 s? Correct answer: 0 . 401674 m / s 2 (tolerance 1 %). Explanation: Basic Concepts = t a t = r v t = r = + t The final angular velocity is 65 . 7 rpm = 6 . 88009 rad / s, and the radius of the record is equal to 15 . 4 in ./ 2 = 0 . 19558 m. = t = 6 . 88009 rad / s 3 . 35 s = 2 . 05376 rad / s 2 . So a t = r = (0 . 19558 m)(2 . 05376 rad / s 2 ) = 0 . 401674 m / s 2 ....
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