##### A light rod of length ℓ = 1.00 m rotates about an axis perpendicular to its leng

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Jun 11th, 2015

The kinetic energy of a mass M rotating at distance R from an axis with angular velocity w , is

K = (1/2)Iw^2

where I is the mass's moment of inertia, which is just MR^2 if its concentrated at distance R.

So for part (a)

K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2

=(1/2)[M1 + M2]R^2w^2

= (1/2)[4.65 + 3](.5)^2(2.2)^2

= 4.63 J

In part (b) you have to add the moment of inertia of the rod which is I = (1/12)ML^2 = (1/12)(2.05)(1)^2 = .171

K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2 + (1/2)Iw^2

= 4.63 + (1/2)(.171)(2.2)^2

= 5.04 J

Jun 11th, 2015

hi this is mjoody

Jun 11th, 2015

hi this is mjoody

Jun 11th, 2015

hi this is mjoody

Jun 11th, 2015

hi this is mjoody

Jun 11th, 2015

hi this is mjoody

Jun 11th, 2015

hi this is mjoody

Jun 11th, 2015

sorry not me the computer

Jun 11th, 2015

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Jun 11th, 2015
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Jun 11th, 2015
Oct 21st, 2017
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