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

A tower clock has an hour hand 2.70 m long with a mass of 60.o kg, and a minute hand 4.50 m long with a mass of 100 kg. Calculate the total rotational kinetic energy of the two hands about the axis of rotation.

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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