Calculating Rotational Kinetic Energy of a Clock's Hands

[elkedoring]

[SOLVED] Energy in Rotational Motion

There is a clock that has an hour hand 2.7m long with a mass of 60 kg. The minute hand is 4.5m log with a mass of 100 kg. What is the total rotational kinetic energy assuming the hands act as long thin rods?



K=I*rotation vel.^2
I=(1/3) ML^2 ( I think this is the right equation to use)


Ok so first I found the rotational velocities. The minute hand move 1 rev./60min and the hour hand moves 1rev/12hrs. Next I converted them into rads/sec. giving a rotational velocity of .000145444 rads/sec for the first one and .001745329 rads/sec on the second.
Then I went back and plugged in all the numbers for the K equation. [(1/3)*60*2.7^2*.000145444] + [(1/3)*100*4.5^2*.001745329]. This gave me an answer of .00206. This answer is about double what the answer should be (.00103). The only place I thik I may have gone wrong is at the (1/3). Should it (1/6) because the rods are pivoting in between the center of mass (which would use (1/12)) and the end (which is the (1/3) that I used)? Could someone please help me see where I went wrong?

 

[G01]

At first glance, I noticed you did not square your rotational velocities.

 

[Nabeshin]

Rotational kinetic energy is $$\frac{1}{2}$$*$$I$$*$$\omega^{2}$$, so I think that's where you're getting twice the intended answer.

 

[elkedoring]

Thank you! I hadn't taken the 1/2 into consideration! That solves it!