12_P68InstructorSolution

# 12_P68InstructorSolution - 12.68 Model The bar is a solid...

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12.68. Model: The bar is a solid body rotating through its center. Visualize: Solve: (a) The two forces form a couple. The net torque on the bar about its center is net I LF I F L α τα == = where F is the force produced by one of the air jets. We can find I and as follows: 22 2 2 10 1 0 11 (0.50 kg)(0.60 m) 0.015 kg m 12 12 ( ) 150 rpm 5.0 rad/s 0 rad (10 s 0 s) 0.50 rad/s (0.015 kg m )(0.5 rad/s ) 0.0393 N (0.60 m) IM L tt F ωωα π = =+ −⇒ = = + − ⇒= ⇒= = The force 39 mN. F = (b) The torque of a couple is the same about any point. It is still net . LF τ = However, the moment of inertia has changed. 2
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## This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.

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