A hollow cylinder closed at both ends has an outside diameter of 1.25 m, length 3.5 m and specific weight 75 KN/`m^3`. If cylinder is to float just in stable equilibrium in sea water ( specific weight 10 KN/`m^3`), find its minimum permissible thickness.

SOLUTION:

Let,

h be the depth of immersion and t be the minimum possible thickness

Now, from the law of floatation,

Weight of cylinder = weight of water displaced 

`V_(cyl i nder) **sp.weight=pi/4**1.25**h**10`

or,`(Volume\ of\ two\ end\ Section + Volume\ of\ Circ u lar Po rtion)*Sp.Wt.=pi/4**15.625h `

`{2**pi/4d^2t+pi/4 (d^2-(d-2t)^2)L}w_c=12.2718h`

Assuming `t < < L`

`{2**pi/4d^2t+pi/4 (d^2-(d^2-4dt+4t^2))L}w_c=12.2718h`

`{(pid^2t)/2+p....Show More

SOLUTION:

Let,

h be the depth of immersion and t be the minimum possible thickness

Now, from the law of floatation,

Weight of cylinder = weight of water displaced 

`V_(cyl i nder) **sp.weight=pi/4**1.25**h**10`

or,`(Volume\ of\ two\ end\ Section + Volume\ of\ Circ u lar Po rtion)*Sp.Wt.=pi/4**15.625h `

`{2**pi/4d^2t+pi/4 (d^2-(d-2t)^2)L}w_c=12.2718h`

Assuming `t < < L`

`{2**pi/4d^2t+pi/4 (d^2-(d^2-4dt+4t^2))L}w_c=12.2718h`

`{(pid^2t)/2+pi/4 (4dt)L}w_c=12.2718h`

`{(pi**1.25^2t)/2+pi**1.25tL}7.5=12.2718h`

`h=99t`

Now, Volume of seawater displaced is,

`V=((pi/2**1.25^2t+pi**1.25**t**3.5)7.5)/10`

`=121.5t`

Let M be the point ot metacentre.

Thus,

`BM=I_(yy)/V`

`=pi/64**1.25^4/(121.5t)`

`=0.00099/t`

`OB=h/2=49.5t`

`OG=3.5/2=1.75 m`

Now,

`BG=OG-OB`

`=1.75-49.5t`

Now for the cylinder to just float in stable equilibrium,

`GM=0`

or,`BM-BG=0`

`1.75-49.5t=0.00099/t`

Solving this we get two values of t as,

`t=0.03478` and `t=5.75**10^(-4)`

Clearly,

Minimum permissible thickness is, `t=5.75**10^(-4) m`