1.

A uniform circular disc of radius R oscillates in a vertical plane about a horizontal axis. Find the distance of the axis of rotation from the centre of which the period is minimum. What is the value of this period?

Answer» 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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