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The middle of a uniform rod of mass m and length l is rigidly fixed to a vertical axis OO^' so that the angle between the rod and the axis is equal to theta (figure). The ends of the axis OO^' are provided with bearings. The system rotates without friction with an angular velocity omega. Find: (a) the magnitude and direction of the rod's angular momentum M relative to the point C, as well as its angular momentum relative to the rotation axis, (b) how much the modulus of the vector M relative to the point C increases during a half-turn, (c) the moment of external forces N acting on the axle OO^' in the process of rotation. |
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Answer» Solution :(a) The ANGULAR velocity `omega` about `OO^'` can be resolved into a component parallel to the rod and a component `omega sin theta` PERPENDICULAR to the rod through C. The component parallel to the rod does not contribute so the angular momentum `M=lomegasintheta=(1)/(12)ml^2omegasintheta` ALSO, `M_z=Msintheta=(1)/(12)ml^2omegasin^2theta` This can be OBTAINED directly also, (b) The modulus of `vecM` does not change but the modulus of the change of `vecM` is `|DeltavecM|`. `|DeltavecM|=2Msin(90-theta)=(1)/(12)ml^2omegasin2theta` (c) Here `M_(_|_)=Mcostheta=Iomegasinthetacostheta` Now `|(dvecM)/(dt)|=Iomegasinthetacostheta(omegadt)/(dt)=(1)/(24)ml^2omega^2sin^2theta` as `vecM` precesses with angular velocity `omega`. 
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