A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for its to have minimum time period ?

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Solution

Let mass of uniform disc, = m

Then minimum time period,

$T = 2 π \sqrt{\frac{I}{m g . \sqrt{2} r}}$

$= 2 π \sqrt{\frac{2 m r^{2}}{\sqrt{2} m g r}}$

$= 2 π \sqrt{\frac{\sqrt{2} r^{2}}{g}}$

Clearly, for minimum time period distance of hole from the centre $= \frac{r}{\sqrt{2}} .$

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A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for its to have minimum time period ?

Let mass of uniform disc, = m Then minimum time period, T=2π √Img. √2 r =2π √2mr2√2 mgr =2π √√2 r2g Clearly, for minimum time period distance of hole from the centre =r√2.

Let mass of uniform disc, = m

Then minimum time period,

$T = 2 π \sqrt{\frac{I}{m g . \sqrt{2} r}}$

$= 2 π \sqrt{\frac{2 m r^{2}}{\sqrt{2} m g r}}$

$= 2 π \sqrt{\frac{\sqrt{2} r^{2}}{g}}$

Clearly, for minimum time period distance of hole from the centre $= \frac{r}{\sqrt{2}} .$