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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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