Fluid Mechanics Concept Page - 6

Example

Torque due to buoyant force

Example:
A slender homogeneous rod of length 2L$2L$ floats partly immersed in water, being supported by a string fastened to one of its end, as shown. The specific gravity of the rod is 0.75$0.75$. Find the length of rod that extends out of water.Solution:
Lets say x$x$ length of the rod is dipped into the water. 
Since the buoyant force acts through the center of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point O$O$ along the line of action of T,$T,$
∑τo=0$\sum {\tau }_o=0$
⇒0=wlcosθ−FB(2l−x2)cosθ$\Rightarrow 0=wl\cos \theta -F_B(2l-{\displaystyle \frac{x}{2}})\cos \theta$
⇒0=ρrodgA(2l)(lcosθ)−ρwatergAx(2l−x2)cosθ$\Rightarrow 0={\rho }_{rod}gA(2l)(l\cos \theta )-{\rho }_{water}gAx(2l-\frac{x}{2})\cos \theta$
⇒0=(12ρwatergAcosθ)$\Rightarrow 0=(\frac{1}{2}{\rho }_{water}gA\cos \theta )$ (x2−4lx+4ρrodρwaterl2)$(x^2-4lx+4\frac{{\rho }_{rod}}{{\rho }_{water}}{l}^2)$ where A=cross section area$A=\text{cross section area}$
⇒x2−4lx+3l2=0$\Rightarrow x^2-4lx+3l^2=0$
⇒x=l,3l$\Rightarrow x=l,3l$. 
x=3l$x=3l$ is not a possible solution, Hence 2l−x=2l−l=l$2l-x=2l-l=l$ length of the rod extends out of water.

Definition

Apparent weight of floating body

Apparent weight is given by subtracting the upthrust on the body from the weight of the body. 
Case 1- Floating body: Wapparent=0$W_{apparent}=0$ because weight equals upthrust
Case 2- Sinking body: Wapparent=(ρ−ρo)Vg$W_{apparent}=(\rho -{\rho }_o)Vg$
                      where ρ:$\rho :$ density of the body 
                                 ρo:${\rho }_o:$ density of the fluid
                                 V:$V:$ volume of the body

Example

Buoyant Force on an immersed object in terms of specific gravity

Example: When a body lighter than water is completely submerged in water, the
buoyant force acting on it is found to be n times its weight. Find the specific gravity of the material of the body.

Solution:Let volume of the body be V$V$ and its density be ρs${\rho }_s$ 
Let density of fluid be ρ$\rho$
Weight of the body is Vρsg$V{\rho }_{s}g$
Weight of the fluid displaced = Buyoant Force = Vρg$V\rho g$
n$n$(weight of body) =$=$ Buoyant force acting
⇒n(Vρsg)=(Vρg)$\Rightarrow n(V{\rho }_{s}g)=(V\rho g)$
ρsρ=1n${\displaystyle \frac{{\rho }_s}{\rho }}={\displaystyle \frac{1}{n}}$.

Definition

Centre of Buoyancy

The centre of gravity of the volume of displaced fluid when an object is immersed in the fluid is called the centre of buoyancy.

Definition

Center of Gravity and Center of Buoyancy

• Center of Gravity is the point in a body where the gravitational force may be taken to act.
• Center of Buoyancy is the center of the gravity of the volume of water which a body displaces.

Example

Torque due to buoyant force when centre of buoyancy is different from the center of mass

Example:
A non-uniform cylinder of mass m, length l and radius r is having its centre of mass at a distance l/4 from the centre and lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density ρ$\rho$. The moment of inertia of the rod about the centre of mass is l. Find the angular acceleration of point A relative to point B just after the rod is released from the position as shown in the figure.
Solution:Buoyancy force will act at center and Force of gravity will act at center of mass.
Balancing torques, πr2lρg×l/4=Iα$\pi r^{2}l\rho g\times l/4=I\alpha$⇒α=πr2l2ρg/4I

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