This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Ph1a Solutions 6 Chefung Chan (cchan@caltech.edu), Fall 2004 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 6.1 Serway 10.23(6th)/10.28(5th) (5 points) For rotation about the center of mass, only the rods perpendicular to the axis of rotation contribute to the moment of inertia. I cm = 2 parenleftbigg 1 12 mL 2 parenrightbigg = 1 6 mL 2 Then apply the parallel axis theorem to shift the rotation axis to the end of one of the rods: I = I cm + (3 m ) parenleftbigg L 2 parenrightbigg 2 = 11 12 mL 2 6.2 Serway 10.66(6th)/10.64(5th) (5 points) Note: The numerical values for this problem may vary slightly depending on the values used for the length in seconds of 1 day and 1 year (the calendar values or the astronomical values). First part (3 points) The moment of inertia of the Earth is I = 2 5 MR 2 = (0 . 4)(5 . 98 10 24 kg)(6 . 37 10 6 m) 2 = 9 . 71 10 37 kg m 2 The rotational velocity is = 2 radians 1 day = 2 / (86 400 s) = 7 . 27 10 5 s 1 The resulting kinetic energy is K = 1 2 I 2 = (0 . 5)(9 . 71 10 37 kg m...
View
Full
Document
This note was uploaded on 03/30/2010 for the course PH 1a taught by Professor Goodstein during the Fall '07 term at Caltech.
 Fall '07
 Goodstein
 Work

Click to edit the document details