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HW10 - homework 10 GAUTHIER JOSEPH Due Nov 9 2007 1:00 am...

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homework 10 – GAUTHIER, JOSEPH – Due: Nov 9 2007, 1:00 am 1 Question 1, chap 11, sect 3. part 1 of 1 10 points A uniform disk of radius 1 . 7 m and mass 5 . 6 kg is suspended from a pivot 0 . 323 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 . 42936 rad / s (tolerance ± 1 %). Explanation: Basic Concepts The physical pendulum: τ = I α = m g d sin θ α = d 2 θ dt 2 so that the angular frequency for small oscil- lations (sin θ θ ) is ω = radicalbigg m g d I . Parallel axis theorem I = I 0 + m a 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 m R 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 m R 2 + m d 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 m g d I P = radicaltp radicalvertex radicalvertex radicalvertex radicalbt m g 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 ) (0 . 323 m) (1 . 7 m) 2 2 + (0 . 323 m) 2 = 1 . 42936 rad / s . Question 2, chap 11, sect 3. part 1 of 3 10 points Hint: The moment of inertia of a uniform rod about its center-of-mass is 1 12 M L 2 . Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal bearing at a point O parenleftbigg 5 8 L from the lower end parenrightbigg , as shown in the figure. 5 8 L L O θ The moment of inertia I of the rod about the pivot point O is given by
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homework 10 – GAUTHIER, JOSEPH – Due: Nov 9 2007, 1:00 am 2 1. I = 13 81 M L 2 2. I = 13 75 M L 2 3. I = 19 147 M L 2 4. I = 19 81 M L 2 5. I = 19 192 M L 2 correct 6. I = 31 147 M L 2 7. I = 43 192 M L 2 8. I = 7 81 M L 2 9. I = 13 147 M L 2 10. I = 1 9 M L 2 Explanation: Basic Concepts: Using the parallel axis theorem, the momentum of inertia is I M d 2 . In this case there are two masses with I O = I cm + M D 2 , where D = parenleftbigg 5 8 1 2 parenrightbigg L = 1 8 L , so (1) I O = M bracketleftBigg 1 12 + parenleftbigg 1 8 parenrightbigg 2 bracketrightBigg L 2 = 19 192 M L 2 . (2) Question 3, chap 11, sect 3. part 2 of 3 10 points Hint: Use the small angle approximation. If θ is in radians, then Newton’s second law for rotational motion for this pendulum is given by 1. d 2 θ dt 2 = 45 26 g L θ 2. d 2 θ dt 2 = 9 14 g L θ 3. d 2 θ dt 2 = 105 62 g L θ 4. d 2 θ dt 2 = 3 2 g L θ 5. d 2 θ dt 2 = 12 7 g L θ 6. d 2 θ dt 2 = 63 38 g L θ 7. d 2 θ dt 2 = 21 26 g L θ 8. d 2 θ dt 2 = 24 19 g L θ correct 9.
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