A solid cone floats in water with its apex downwards. Determine least apex angle of cone for stable equilibrium. The specific gravity of cone is 0.8.

SOLUTION:

Let,

D be the diameter of cone

d be the diameter of cone at water level

`2theta` be the apex angle of the cone

H be the height of cone

h be the depth of cone in water

G be the c.g. of the cone

B be the C.B. of the cone

We know that,  for cone,

`AG=3/4H` and `AB=3/4h`

Also,

Volume of water displaced is,

`=1/3pir^2h`

And the volume of the cone is,

`=1/3piR^2H`

Thus the weight of the cone is,

`=800**9.81**1/3**piR^2H`

Now from `triangle AEF`,

`tantheta=R/H`

`R=H tantheta`

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SOLUTION:

Let,

D be the diameter of cone

d be the diameter of cone at water level

`2theta` be the apex angle of the cone

H be the height of cone

h be the depth of cone in water

G be the c.g. of the cone

B be the C.B. of the cone

We know that,  for cone,

`AG=3/4H` and `AB=3/4h`

Also,

Volume of water displaced is,

`=1/3pir^2h`

And the volume of the cone is,

`=1/3piR^2H`

Thus the weight of the cone is,

`=800**9.81**1/3**piR^2H`

Now from `triangle AEF`,

`tantheta=R/H`

`R=H tantheta`

Similarly, `r=htantheta`

Thus the weight of cone is,

`=800g**1/3pi**(Htantheta)^2**H`

`=(800g**pi**H^3tan^2theta)/3`

Weight of water displaced is given as,

`=(1000g**pi**h^3tan^2theta)/3`

Now from the law of floatation,

Weight of cone = weight of water displaced 

`(800g**pi**H^3tan^2theta)/3=(1000g**pi**h^3tan^2theta )/3`

`800H^3=100h^3`

`H/h=(10/8)^(â??)`

Also the metacentric height is,

`GM=I_(yy)/V-BG`

Where, `I_(yy)`=M.I of the plan at water line about y axis.

`=(pid^4)/64`

And V=volume of cone in water

`=1/3pir^2*h`

`=â??(pid^2h)/4`

And `BG=3/4 (H-h)`

Thus,

`GM=((pid^4)/64)/((pid^2h)/12)-3/4 (H-h)`

`=(12r^2)/(16h)-3/4(H-h)`

`=3/4htan^2theta-3/4 (H-h)`

For stability,

`GM >0`

`htan^2theta>H-h`

`1+tan^2theta>(10/8)^(â??)`

`theta>15^0 31'`

`2theta >31^0 3'`

Thus the minimum apex angle is `31^0 3'`.