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

$${A}\:{disc}\:{of}\:{radius}\:{r}\:{suspended}\:{from} \\ $$$${a}\:{point}\:{lie}\:{on}\:{itself}.{Find}\:{out}\:{the} \\ $$$${minimum}\:{time}\:{period}\:{of}\:{oscillation} \\ $$$${of}\:{the}\:{disc}. \\ $$

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