{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw25 - Muraj Hamza Homework 25 Due 4:00 am Inst Florin This...

This preview shows pages 1–2. Sign up to view the full content.

Muraj, Hamza – Homework 25 – Due: Mar 31 2006, 4:00 am – Inst: Florin 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A solid sphere of radius 21 cm is positioned at the top of an incline that makes 28 angle with the horizontal. This initial position of the sphere is a vertical distance 2 . 5 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. The acceleration of gravity is 9 . 8 m / s 2 . Hint: The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 . 21 cm M μ 28 2 . 5 m Calculate the speed of the sphere when it reaches the bottom of the incline in the case where it rolls without slipping. Correct answer: 5 . 91608 m / s. Explanation: Basic Concepts: X E = K trans + K rot + U gravity = const From conservation of energy we have M g h = 1 2 M v 2 1 + 1 2 I ω 2 = 1 2 M v 2 1 + 1 2 2 5 M R 2 v 2 1 R 2 = 7 10 M v 2 1 . Therefore, v 1 = r 10 7 g h = r 10 7 (9 . 8 m / s 2 ) (2 . 5 m) = 5 . 91608 m / s . 002 (part 2 of 2) 10 points Calculate the speed of the sphere when it reaches the bottom of the incline in the case where it slips frictionlessly without rolling. Correct answer: 7 m / s. Explanation: From conservation of energy we have M g h = 1 2 M v 2 2 . Therefore, v 2 = p 2 g h = q 2 (9 . 8 m / s 2 ) (2 . 5 m) = 7 m / s . 003 (part 1 of 3) 10 points Consider a thick cylindrical shell rolling down along an incline. It has a mass m , an outer ra- dius R and a moment of inertia I cm = 4 5 mR 2 about its center of mass. The incline is at an angle θ to the horizontal. Its length is s and its height is h . s h θ v A A The equation of motion τ = as the shell is rolling down the incline is given by(using the contact point as the pivot point): 1. mgR = I cm α 2. mgR sin θ = ( I cm + mR 2 ) α correct 3. mgR tan θ = I cm α 4. mgR = ( I cm + mR 2 ) α 5. mgR sin θ = I cm α

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}