InterviewSolution
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InterviewSolution
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A cylindrical disc of a gyroscope of mass m=15kg and radius r=5.0cm spins with an angular velocity omega=330rad//s. The distance between the bearings in which the axle of the disc is mounted is equal to l=15cm. The axle is forced to oscillate about a horizontal axis with a period T=1.0s and amplitude varphi_m=20^@. Find the maximum value of the gyropscopic forces exerted by the axle on the bearings. |
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Answer» Solution :The MOMENT of inertia is `1/2mr^2` and angular momentum is `1/2mr^2omega`. The axis oscillates about a horizontal axis making an INSTANTANEOUS ANGLE. `varphi=varphi_m"sin"(2pit)/(T)` This MEANS that there is a variable precession with a rate of precession `(dvarphi)/(dt)`. The MAXIMUM value of this is `(2pivarphi_m)/(T)`. When the angle between the axle and the axis is at the maximum value, a torque `IomegaOmega` `=1/2mr^2omega(2pivarphi_m)/(T)=(pimr^2omegavarphi_m)/(T)` acts on it. The corresponding gyroscopic force will be `(pimr^2omegavarphi_m)/(lT)=90N` |
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