PHY 303K - Homework 17 - homework 17 FIERRO, JEFFREY Due:...

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Unformatted text preview: homework 17 FIERRO, JEFFREY Due: Mar 29 2008, 11:00 pm 1 Question 1, chap 12, sect 5. part 1 of 1 10 points An 853 kg car is held in place by an un- realiable winch. The gearbox of the winch breaks and at that moment the car-drum sys- tem free-falls from rest, shown in the figure. During the cars fall, there is no slipping be- tween the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 171 kg m 2 and that of the pulley is 7 kg m 2 . The radius of the drum is 22 cm and the radius of the pulley is 9 cm . The acceleration due to gravity is 9 . 81 m / s 2 . 853 kg 22 cm 171 kg m 2 7 kg m 2 7 m Find the speed of the car as it hits the ground. Correct answer: 4 . 72371 m / s (tolerance 1 %). Explanation: Let : I d = 171 kg m 2 , I p = 7 kg m 2 , r d = 22 cm = 0 . 22 m , r p = 9 cm = 0 . 09 m , m b = 853 kg , and h = 7 m . Applying conservation of mechanical en- ergy ( K i = U f = 0) and v = r , 1 2 mv 2 + 1 2 I d 2 d + 1 2 I p 2 p mg h = 0 1 2 mv 2 + 1 2 I d v 2 r 2 d + 1 2 I p v 2 r 2 p mg h = 0 v = radicaltp radicalvertex radicalvertex radicalvertex radicalbt 2 mg h m + I d r 2 d + I p r 2 p = radicaltp radicalvertex radicalvertex radicalvertex radicalbt 2 (853 kg) (9 . 81 m / s 2 ) (7 m) (853 kg) + (171 kg m 2 ) (0 . 22 m) 2 + (7 kg m 2 ) (0 . 09 m) 2 = 4 . 72371 m / s . Question 2, chap 12, sect 5. part 1 of 1 10 points A massless rod of length L has a mass m fastened at its center and another mass m fastened at one end. On the opposite end, the rod is pivoted with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down, as shown in the figure below. L L / 2 m radius R L m What is the angular velocity as the rod swings through its lowest (vertical) position? 1. radicalbigg 2 g 3 L 2. radicalbigg 6 g 5 L 3. radicalbigg 3 g 2 L homework 17 FIERRO, JEFFREY Due: Mar 29 2008, 11:00 pm 2 4. radicalbigg 5 g 6 L 5. radicalbigg 4 g 5 L 6. radicalbigg 12 g 5 L correct 7. radicalbigg g 2 L 8. radicalbigg g L 9. radicalbigg g 4 L 10. radicalbigg 2 g L Explanation: The mechanical energy of the system is con- served. Measuring heights from the point at the bottom of the rod when it is vertical, the initial potential energy of the system is U i = (2 m ) g L . The potential energy at the bottom of the swing, U f = mg parenleftbigg L 2 parenrightbigg , so U = parenleftbigg 2 1 2 parenrightbigg mg L = 3 2 mg L . The moment of inertia of the system is I = m parenleftbigg L 2 parenrightbigg 2 + mL 2 = 5 4 mL 2 . So the change in the kinetic energy is K = 1 2 I 2 = 5 8 mL 2 2 Therefore, U = K 3 2 mg L = 5 8 mL 2 2 12 g = 5 L 2 , so = radicalbigg 12 g 5 L ....
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This homework help was uploaded on 04/03/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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PHY 303K - Homework 17 - homework 17 FIERRO, JEFFREY Due:...

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