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November 26, 2014

November 26, 2014

Posted by **Jc** on Sunday, October 23, 2011 at 8:32pm.

- Physics -
**Kayrie**, Wednesday, March 13, 2013 at 8:13pmUse the rotational energy equation, E = 0.5Iω², where I is the moment of inertia for a long thin rod with an axis through it's midpoint (I = (1/12)mL²).

E = 0.5Iω²

E = 0.5 [1/12)mL²] ω²

E = (1/24) mL² ω²

Find ω by multiplying the rpm by 2π and then dividing by 60 to put units into rad/sec (or multiply ω by π/30)

3450 * 2π = 21 676.9... rad/min

(21 676.9... rad/min)/60 = 361.28... rad/sec

Insert into energy eqn and solve

E = (1/24) mL² (361.28...rad/s)²

E = (1/24)(0.65kg)(0.55m²)(361.28...rad/s)²

E = ~1069 J

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