ps9b23 - Ph 111.01 October 29, 2008 PS 9b (24) Problem 1. A...

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Ph 111.01 October 29, 2008 PS 9b (24) Problem 1. A 2 kg mass and a 5 kg mass are attached to either end of a 3 m long massless rod. a.) Find the center of mass of the system. m, from the 2 kg mass. Find the rotational inertia (I) of the system when rotated about: b.) the end with the 2 kg mass. kg m 2 c.) the end with the 5 kg mass. kg m 2 d.) the center of the rod. kg m 2 e.) the center of mass of the system. kg m 2 (Compare this to parts b-d. Is this what you expect?) f.) If the system is rotated about the center of mass by a force of 8 N acting on the 5 kg mass, (perpendicular to the rod), what will the size of the angular acceleration of the system be? rad/s 2 m 1 m 2 x 0L Solution. (a) We will calculate the x -coordinate of the CM, with x = 0 at the position of the smaller mass, as shown. () ( ) ( ) 11 2 2 12 2 kg 0 5 kg 3 m 2.143 m 7 kg CM Xm x m x mm =+ = + = + (b) To find the rotational inertia (the moment of inertia I ) about various points, we will use the simple expression for point masses, 2 I MR = . ( )( ) 2 22 2 5 kg 3 m 45 kg-m Im L == = (c) Now about the big mass: ( )( ) 2 1 2 kg 3 m 18 kg-m L = (d) Now about the point x = L /2: ( ) ( ) 2 2 kg 1.5 m 5 kg 1.5 m 15.75 kg-m LL m ⎛⎞ =+= + = ⎜⎟ ⎝⎠ (e) About x = 2.143 m: ( ) ( )( ) ( )( ) 2222 2 2.143 m 2.143 m 2 kg 2.143 m 5 kg 0.857 m 12.86 kg-m mL = + = It is interesting that the minimum moment of inertia is about the CM.
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This note was uploaded on 05/13/2010 for the course PHYSICS 111 taught by Professor Staff during the Fall '08 term at S.F. State.

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ps9b23 - Ph 111.01 October 29, 2008 PS 9b (24) Problem 1. A...

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