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November 27, 2014

November 27, 2014

Posted by **Anderson** on Wednesday, April 18, 2007 at 6:51pm.

Sol: I know that Kr = 1/2Iw^2 or moment of inertia * omega squared. The problem is, it seems like I'm missing omega. I cannot seem to figure out how to get another angular (kinematics) into this problem. Perhaps there are too many unknown that I'm not aware of. Help would be appreciated. thx

teacher's answer: 1.04 E -3 J

Omega for the minute hand is 2PI/minute

Omega for the hour hand is 2PI/hour.

change both to radians per second.

Total rotational energy = 1/2 Ihour*wh^2 + 1/2 Imin*wm^2

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