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HW_06_S

HW_06_S - Ph1a Solutions 6 Chefung Chan([email protected]

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Unformatted text preview: Ph1a Solutions 6 Chefung Chan ([email protected]), 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 ω = 2 π radians 1 day = 2 π/ (86 400 s) = 7 . 27 × 10 − 5 s − 1 The resulting kinetic energy is K = 1 2 Iω 2 = (0 . 5)(9 . 71 × 10 37 kg m...
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HW_06_S - Ph1a Solutions 6 Chefung Chan([email protected]

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