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An uniform disc is rotating at a constan...

An uniform disc is rotating at a constant speed in a vertical plane about a fixed horizontal axis passing through the centre of the disc. A piece of the disc from its rim detaches itself from the disc at the instant when it is at horizontal level with the centre of the disc and moving upwards, then about the fixed axis.
STATEMENT-1 : Angular speed of the disc about the aixs of rotation will increase.
and
STATEMENT-2 : Moment of inertia of the disc is decreased about the axis of rotation.

Text Solution

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The time period of a compound pendulum is the minimum when its length is equal to the radius of gyration about its centre of gravity , i.e., l=k.
Since, the moment of inertia of a disc about an axis perpendicular to its plane and passing through its centre is equal to,
`I=MK^(2)=(1)/(2)MR^(2)implies K=(R)/(sqrt(2))`
Thus, the disc oscillate with the minimum time period when the distance of the axis of rotation from the centre is `(R)/(sqrt(2))` . And the value of this minimum time period will be
`T_("min")=2pisqrt((2R//sqrt(2))/(g)) =2pisqrt((sqrt(2)R)/(g)` or `T_("min")~~2pisqrt((1.414R)/(g))`
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