This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points There is a moon orbiting an Earthlike planet. The mass of the moon is 5 . 03 × 10 22 kg, the centertocenter separation of the planet and the moon is 7 . 48 × 10 5 km, the orbital period of the moon is 25 . 1 days, and the radius of the moon is 1520 km. What is the angular momentum of the moon about the planet? Correct answer: 8 . 15387 × 10 34 kg m 2 / s. Explanation: The angular speed ω in radians per unit time, for a complete circle is ω = 2 π T . Angular momentum L is L = mv r = mr 2 ω = 2 π mr 2 T = 2 π (5 . 03 × 10 22 kg) × bracketleftbigg (7 . 48 × 10 5 km) 1000 m 1 km bracketrightbigg 2 (25 . 1 days) parenleftbigg 24 hr 1 day parenrightbigg = 8 . 15387 × 10 34 kg m 2 / s . 002 (part 1 of 3) 10.0 points A 3 kg bicycle wheel rotating at a 2753 rev / min angular velocity has its shaft supported on one side, as shown in the fig ure. The wheel is a hoop of radius 0 . 6 m, and its shaft is horizontal. The distance from the center of the wheel to the pivot point is 0 . 4 m. When viewing from the left (from the posi tive xaxes), one sees that the wheel is rotat ing in a clockwise manner. Assume: All of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9 . 8 m / s 2 . y z x b mg ω R The magnitude of the angular momentum of the wheel is given by 1. bardbl vector L bardbl = mR 2 ω correct 2. bardbl vector L bardbl = 1 4 mR 2 ω 3. bardbl vector L bardbl = mR 2 ω 2 4. bardbl vector L bardbl = 1 4 mR 2 ω 2 5. bardbl vector L bardbl = 1 2 mR 2 ω 2 6. bardbl vector L bardbl = 1 2 mR 2 ω Explanation: Basic Concepts: vector τ = d vector L dt Solution: The magnitude of the angular momentum I of the wheel is L = I ω = mR 2 ω , and is along the negative xaxis. 003 (part 2 of 3) 10.0 points Let : m = 3 kg , ω = 2753 rev / min b = 0 . 4 m , and R = 0 . 6 m . Find the change in the precession angle after a 1 . 2 s time interval. Correct answer: 2 . 59689 ◦ . Explanation: homework 29 – Turner – (58185) 1 hinojosa (jlh3938) – homework 29 – Turner – (58185) 2 Let : ω = 2753 rev / min = 2 π (2753 rev / min) (60 s / min) = 288 . 293 rad / s ....
View
Full
Document
This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass, Work

Click to edit the document details