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Energy in Rotational Motion

  1. Apr 13, 2008 #1
    [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?
     
  2. jcsd
  3. Apr 13, 2008 #2

    G01

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    At first glance, I noticed you did not square your rotational velocities.
     
  4. Apr 13, 2008 #3

    Nabeshin

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    Rotational kinetic energy is [tex]\frac{1}{2}[/tex]*[tex]I[/tex]*[tex]\omega^{2}[/tex], so I think that's where you're getting twice the intended answer.
     
  5. Apr 13, 2008 #4
    Thank you! I hadn't taken the 1/2 into consideration! That solves it!
     
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