om/ibiscms/mod/ibis/view.php?id 3390905 y. PHYs 211 Spring17 smNNETT I Activitie

Question

om/ibiscms/mod/ibis/view.php?id 3390905 y. PHYs 211 Spring17 smNNETT I Activities and Due Dates I Hv: Dynamics of Rotational Motion Grade bool 04/26/2017 06:00 PM 46 7/100 4/19/2017 o6:51 PM Print Calculator Periodic Tablo Question 7 of 10 Map A Sapling Learning A trapeze artist performs an aerial maneuver. While in a tucked position, as shown in Figure A, she rotates about her center of mass at a rate of 5.91 rads. Her moment of inertia about this axis is 16.1 kg- m A short time later the aerialist is in the straight position, as shown in Figure B. If the moment of inertia about her center of mass in this position is now 34.9 kg m2, what is her rotational speed? rad/ s Figure A Figure B 19

Explanation / Answer

Initial moment of inertia I1 = 16.1 kg-m2

Initial angular velocity 1 = 5.91 rad/s

Final moment of inertia I2 = 34.9 kg-m2

Let the final angular velocity be 2

Then conservation of anular momentum prnciple says:

I11 = I22

or (16.1 kg-m2)( 5.91 rad/s) = (34.9 kg-m2)2

2 = 2.726 rad/s is her final rotational speed.

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