A gyroscope, a uniform disc of radius R=5.0cm at the end of a rod of length l=10cm (figure), is mounted on the floor of an elevator car going up with a constant acceleration w=2.0m//s^2. The other end of the rod is hinged at the point O. The gyroscope precesses with an angular velocity n=0.5 rps. Neglecting the friction and the mass of the rod, find the proper angular velocity of the disc.

A gyroscope, a uniform disc of radius R=5.0cm at the end of a rod of l

1.	A gyroscope, a uniform disc of radius R=5.0cm at the end of a rod of length l=10cm (figure), is mounted on the floor of an elevator car going up with a constant acceleration w=2.0m//s^2. The other end of the rod is hinged at the point O. The gyroscope precesses with an angular velocity n=0.5 rps. Neglecting the friction and the mass of the rod, find the proper angular velocity of the disc.
Answer» Solution :The moment of inertia of the disc about its symmetry axis is `1/mR^2`. If the angular velocity of the disc is `omega` then the angular MOMENTUM is `1/2mR^2omega`. The precession frequency being `2pin`, we have `\|(dvecM)/(DT)\|=1/2mR^2omegaxx2pin` This MUST equal `m(g+w)l`, the effective gravitational torques (g being REPLACED by `g+w` in the elevator). Thus, `omega=((g+w)l)/(piR^2n)=300rad//s`

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