Homework 16 - homework 16 – FIERRO, JEFFREY – Due: Mar...

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Unformatted text preview: homework 16 – FIERRO, JEFFREY – 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 36 rev/min. The record slows to 15 rev/min in 2.1 s. What is the record’s average angular accel- eration during this time interval? Correct answer: − 1 . 0472 rad / s 2 (tolerance ± 1 %). Explanation: Basic Concept: α avg = ω 2 − ω 1 Δ t Given: ω 1 = 36 rev / min ω 2 = 15 rev / min Δ t = 2 . 1 s Solution: α avg = ω 2 − ω 1 Δ t = 15 rev / min − 36 rev / min 2 . 1 s · 2 π rad 1 rev · 1 min 60 s = − 1 . 0472 rad / s 2 Question 2, chap 11, sect 3. part 1 of 1 0 points A uniform disk of radius 3 . 2 m and mass 6 . 9 kg is suspended from a pivot 2 . 08 m above its center of mass. The acceleration of gravity is 9 . 8 m / s 2 . axis Find the angular frequency ω for small os- cillations. Correct answer: 1 . 46897 rad / s (tolerance ± 1 %). Explanation: Basic Concepts The physical pendulum: τ = I α = − mg d sin θ α = d 2 θ dt 2 so that the angular frequency for small oscil- lations (sin θ ≈ θ ) is ω = 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 ω = radicalBigg mg d I P = radicaltp radicalvertex radicalvertex radicalvertex radicalbt mg d m parenleftbigg R 2 2 + d 2 parenrightbigg or ω = radicaltp radicalvertex radicalvertex radicalvertex radicalbt g d R 2 2 + d 2 = radicaltp radicalvertex radicalvertex radicalvertex radicalbt (9 . 8 m / s 2 ) (2 . 08 m) (3 . 2 m) 2 2 + (2 . 08 m) 2 = 1 . 46897 rad / s . homework 16 – FIERRO, JEFFREY – 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 34 . 2 rpm record of diameter 4 . 07 in . if the record moves from rest to its final angular speed in 4 . 54 s? Correct answer: 0 . 0407753 m / s 2 (tolerance ± 1 %). Explanation: Basic Concepts α = Δ ω Δ t a t = rα v t = rω ω = ω + α t The final angular velocity is 34 . 2 rpm = 3 . 58142 rad / s, and the radius of the record is equal to 4 . 07 in ./ 2 = 0 . 051689 m. α = Δ ω Δ t = 3 . 58142 rad / s 4 . 54 s = 0 . 788859 rad / s 2 . So a t = r α = (0 . 051689 m)(0 . 788859 rad / s 2 ) = 0 . 0407753 m / s 2 ....
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This note was uploaded on 03/25/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Homework 16 - homework 16 – FIERRO, JEFFREY – Due: Mar...

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