A horizontally oriented uniform disc of mass `M` and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which can slide without friction a small body of mass `m`. A light thread running down through the hollow axle of the disc is tied to the body initially the body was located at the edge of the disc and the whole system rotated with and angular velocity `omega_(0)`. Then by means of a force `F` applied to the lower and of the thread the body was slowly pulled to the rotation axis. find: (a). The angular velocity of the system in its final state. (b). The work performed by the force F.

A horizontally oriented uniform disc of mass `M` and radius R rotates

1.	A horizontally oriented uniform disc of mass `M` and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which can slide without friction a small body of mass `m`. A light thread running down through the hollow axle of the disc is tied to the body initially the body was located at the edge of the disc and the whole system rotated with and angular velocity `omega_(0)`. Then by means of a force `F` applied to the lower and of the thread the body was slowly pulled to the rotation axis. find: (a). The angular velocity of the system in its final state. (b). The work performed by the force F.
Answer» (a) As force F on the body is radial so its angular momentum about the axis becomes zero and the angular momentum of the system about the given axis is conserved. Thus `(MR^2)/(2)omega_0+momega_0R^2=(MR^2)/(2)omega` or `omega=omega_0(1+(2m)/(M))` (b) From the equation of the increment of the mechanical energy of the system: `DeltaT=A_(ext)` `1/2(MR^2)/(2)omega^2-1/2((MR^2)/(2)+mR^2)omega_0^2=A_(ext)` Putting the value of `omega` from part (a) and solving we get `A_(ext)=(momega_0^2R^2)/(2)(1+(2m)/(M))`

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