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Engineering and Comp Sci Homework Help
Rigid Body Kinetics: Angular Acceleration of Rod w/ Friction Torque
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[QUOTE="gus_lyon, post: 4535283, member: 490856"] [h2]Homework Statement [/h2] A long slender rod has a mass of 0.66 kg and a length of 648.0 mm. The friction torque at the hinge is 0.41 Nm. What is the angular acceleration of the rod when it is 38 degrees below the horizontal? [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] Torque = moment of inertia x angular acceleration Moment of inertia for a rod = (1/12)*m*(l^2) = (1/12)*0.66*(0.648^2) = 0.0231 Kgm^2 Torque = Radius * Force = (0.648*cos(38))*(9.81*0.66) = 3.306 Nm Angular acceleration = Torque/Moment of inertia = 3.306/0.0231 = 143.12 Which is wrong. [/QUOTE]
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Rigid Body Kinetics: Angular Acceleration of Rod w/ Friction Torque
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