hw34 - Muraj, Hamza Homework 34 Due: Apr 26 2006, 4:00 am...

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

View Full Document Right Arrow Icon
Muraj, Hamza – Homework 34 – Due: Apr 26 2006, 4:00 am – Inst: Florin 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 2) 10 points A man enters a tall tower, needing to know its height. He notes that a long pendulum extends ±rom the ceiling almost to the ²oor and that its period is 13 . 1 s. The acceleration o± gravity is 9 . 8 m / s 2 . How tall is the tower? Correct answer: 42 . 5999 m. Explanation: Given : T = 13 . 1 s . The height o± the tower is approximately the same as the length o± the pendulum. The period o± the pendulum is T = 2 π s L g L = g T 2 4 π 2 = ( 9 . 8 m / s 2 ) (13 . 1 s) 2 4 π 2 = 42 . 5999 m . 002 (part 2 o± 2) 10 points I± this pendulum is taken to the Moon, where the ±ree-±all acceleration is 1 . 67 m / s 2 , what is its period there? Correct answer: 31 . 7341 s. Explanation: Given : a M = 1 . 67 m / s 2 . The period o± the pendulum on the moon is T Moon = 2 π s L a M = 2 π s 42 . 5999 m 1 . 67 m / s 2 = 31 . 7341 s . 003 (part 1 o± 1) 10 points Consider a light rod o± negligible mass and length L pivoted on a ±rictionless horizontal bearing at a point O . Attached to the end o± the rod is a mass M 1 . Also, a second mass M 2 o± equal size ( i.e. , M 1 = M 2 = M ) is attached to the rod µ 2 3 L ±rom the lower end , as shown in the fgure below. 2 3 L M 2 M 1 O θ The period o± this pendulum in the small angle approximation is given by 1. T = 2 π s 65 72 L g 2. T = 2 π s 85 99 L g 3. T = 2 π s 17 20 L g 4. T = 2 π s 29 35 L g 5. T = 2 π s 73 88 L g 6. T = 2 π s 53 63 L g 7. T = 2 π s 5 6 L g correct 8. T = 2 π s 89 104 L g
Background image of page 1

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

View Full DocumentRight Arrow Icon
Muraj, Hamza – Homework 34 – Due: Apr 26 2006, 4:00 am – Inst: Florin 2 9. T = 2 π s 13 15 L g 10. T = 2 π s 65 77 L g Explanation: Basic Concepts: The momentum of iner- tia is I M d 2 . In this case there are two masses with
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

hw34 - Muraj, Hamza Homework 34 Due: Apr 26 2006, 4:00 am...

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