PHY 303K - Homework 18 - homework 18 FIERRO, JEFFREY Due:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 18 FIERRO, JEFFREY Due: Mar 31 2008, 11:00 pm 1 Question 1, chap 12, sect 5. part 1 of 1 10 points Calculate the moment of inertia for b b b b b Axis 4 m 3 m m 2 m L L L L L Rods of length L are massless. 1. 41 mL 2 2. 30 mL 2 3. 16 mL 2 4. 52 mL 2 5. 32 mL 2 correct Explanation: mL 2 + 2 m (2 L ) 2 + 3 m ( (2 L ) 2 + L 2 ) + 4 m ( L 2 + L 2 ) = mL 2 + 8 mL 2 + 3 m (5 L 2 ) + 4 m (2 L 2 ) =(1 + 8 + 15 + 8) mL 2 =32 mL 2 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 L 2. radicalbigg g 2 L 3. radicalbigg 4 g 5 L 4. radicalbigg g L 5. radicalbigg 12 g 5 L correct 6. radicalbigg 2 g 3 L 7. radicalbigg 3 g 2 L 8. radicalbigg 6 g 5 L 9. radicalbigg 5 g 6 L 10. radicalbigg g 4 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 homework 18 FIERRO, JEFFREY Due: Mar 31 2008, 11:00 pm 2 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 . Question 3, chap 12, sect 5. part 1 of 2 10 points A wheel is formed from a hoop and three equally spaced spokes. The hoops radius is the length of each spoke. . 13 kg 1 . 1 kg . 57 m There are three equally spaced spokes. Each spoke has mass 0 . 13 kg. The hoops mass is 1 . 1 kg. The hoops radius is 0 . 57 m. Determine the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel. Correct answer: 0 . 399627 kg m 2 (tolerance 1 %). Explanation: Let : n = 3 spokes , m = 0 . 13 kg , M = 1 . 1 kg , and R = 0 . 57 m . Each spoke contributes to the total moment of inertia as a thin rod pivoted at one end, hence I = M R 2 + n mR 2 3 = bracketleftBig M + n m 3 bracketrightBig R 2 = bracketleftbigg (1 . 1 kg) + 3 (0 . 13 kg) 3 bracketrightbigg (0 . 57 m) 2 = 0 . 399627 kg m 2 . Question 4, chap 12, sect 5. part 2 of 2 10 points Determine the moment of inertia of the wheel about an axis through its rim and per- pendicular to the plane of the wheel....
View Full Document

Page1 / 13

PHY 303K - Homework 18 - homework 18 FIERRO, JEFFREY Due:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online