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

Physics
Tutor: None Selected Time limit: 1 Day

Jun 11th, 2015

Thank you for the opportunity to help you with your question!

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

Please let me know if you need any clarification. I'm always happy to answer your questions.
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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