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hw20 - khounvivongsy(sk27799 – Homework 20 – Weathers...

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Unformatted text preview: khounvivongsy (sk27799) – Homework 20 – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniform disk of radius 3 . 9 m and mass 1 . 1 kg is suspended from a pivot 1 . 17 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 . 13036 rad / s. 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 ) (1 . 17 m) (3 . 9 m) 2 2 + (1 . 17 m) 2 = 1 . 13036 rad...
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hw20 - khounvivongsy(sk27799 – Homework 20 – Weathers...

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