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A-disc-of-radius-r-suspended-from-a-point-lie-on-itself-Find-out-the-minimum-time-period-of-oscillation-of-the-disc-




Question Number 53151 by Necxx last updated on 18/Jan/19
A disc of radius r suspended from  a point lie on itself.Find out the  minimum time period of oscillation  of the disc.
$${A}\:{disc}\:{of}\:{radius}\:{r}\:{suspended}\:{from} \\ $$$${a}\:{point}\:{lie}\:{on}\:{itself}.{Find}\:{out}\:{the} \\ $$$${minimum}\:{time}\:{period}\:{of}\:{oscillation} \\ $$$${of}\:{the}\:{disc}. \\ $$
Answered by mr W last updated on 18/Jan/19
i hope i understand your question correctly.  maximum time period:  I=((3Mr^2 )/2)  Iα=−Mgrsin θ≈−Mgrθ  ((3Mr^2 )/2)α=−Mgrθ  α=(d^2 θ/dt^2 )=−((2g)/(3r))θ  ω=(√((2g)/(3r)))  ⇒T=2π(√((3r)/(2g)))
$${i}\:{hope}\:{i}\:{understand}\:{your}\:{question}\:{correctly}. \\ $$$${maximum}\:{time}\:{period}: \\ $$$${I}=\frac{\mathrm{3}{Mr}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}\alpha=−{Mgr}\mathrm{sin}\:\theta\approx−{Mgr}\theta \\ $$$$\frac{\mathrm{3}{Mr}^{\mathrm{2}} }{\mathrm{2}}\alpha=−{Mgr}\theta \\ $$$$\alpha=\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=−\frac{\mathrm{2}{g}}{\mathrm{3}{r}}\theta \\ $$$$\omega=\sqrt{\frac{\mathrm{2}{g}}{\mathrm{3}{r}}} \\ $$$$\Rightarrow{T}=\mathrm{2}\pi\sqrt{\frac{\mathrm{3}{r}}{\mathrm{2}{g}}} \\ $$
Commented by mr W last updated on 18/Jan/19
maybe you meant something different.
$${maybe}\:{you}\:{meant}\:{something}\:{different}. \\ $$
Commented by Necxx last updated on 18/Jan/19
thank you so much.Thats just how  the question was framed
$${thank}\:{you}\:{so}\:{much}.{Thats}\:{just}\:{how} \\ $$$${the}\:{question}\:{was}\:{framed} \\ $$
Commented by mr W last updated on 18/Jan/19
what′s the solution given?
$${what}'{s}\:{the}\:{solution}\:{given}? \\ $$

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