This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 cardrum sys tem freefalls 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 ....
View
Full
Document
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.
 Spring '08
 Turner
 Work

Click to edit the document details