HW24 - homework 24 – PAPAGEORGE MATT – Due 4:00 am 1 Question 1 chap 13 sect 1 part 1 of 2 10 points Consider a circular wheel with a mass m

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 24 – PAPAGEORGE, MATT – Due: Mar 27 2008, 4:00 am 1 Question 1, chap 13, sect 1. part 1 of 2 10 points Consider a circular wheel with a mass m , and a radius R . The moment of inertia about the center of the wheel is I = k mR 2 , where k is a constant in the range between 0 . 5 ≤ k ≤ 1 . . A rope wraps around the wheel. A weight of mass 2 m is attached to the end of this rope. m R 2 m At some moment, the weight is falling with a speed v . The total kinetic energy K of the system at this moment is given by 1. K = [1 + k ] mv 2 2. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 3. K = bracketleftbigg 1 + 2 k 2 bracketrightbigg mv 2 4. K = [1 + k ] mv 5. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 correct 6. K = [1 + 2 k ] mv 7. K = bracketleftbigg 1 + 2 k 2 bracketrightbigg mv 8. K = [1 + 2 k ] mv 2 9. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 10. K = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 Explanation: Basic Concepts: Rotational kinetic en- ergy is E rot = 1 2 I ω 2 K tot = K wheel + K mass At the moment the weight is descending with a speed v , the tangential speed of the wheel is also v . Then the angular velocity of the wheel is ω = v R . So the total kinetic energy is K = 1 2 2 mv 2 + 1 2 I ω 2 = 1 2 2 mv 2 + 1 2 k m ( Rω ) 2 = 1 2 2 mv 2 + 1 2 k mv 2 = bracketleftbigg 1 + k 2 bracketrightbigg mv 2 . Question 2, chap 13, sect 1. part 2 of 2 10 points Assume: k = 1 2 . If the system is released from rest, find the speed v at the moment when the weight has descended a vertical distance h . 1. v = radicalbigg 8 5 g h correct 2. v = radicalbig g h 3. v = radicalbigg 4 5 g h 4. v = radicalbigg 2 3 g h 5. v = radicalbigg 8 9 g h 6. v = radicalbigg 8 11 g h 7. v = radicalbigg 16 11 g h 8. v = radicalbig 2 g h 9. v = radicalbigg 4 3 g h homework 24 – PAPAGEORGE, MATT – Due: Mar 27 2008, 4:00 am 2 10.10....
View Full Document

This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.

Page1 / 5

HW24 - homework 24 – PAPAGEORGE MATT – Due 4:00 am 1 Question 1 chap 13 sect 1 part 1 of 2 10 points Consider a circular wheel with a mass m

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