# forces again

• Jul 1st 2009, 07:07 AM
thereddevils
forces again
When an object moves through a fluid , it experiences a retarding force due to turbulence . For a sphere of radius r moving with a speed of v in a fluid of density , p , the retarding force is given by

$\displaystyle F=kpr^2v^2$

By relating the retarding force to the transfer of momentum between the fluid and the sphere , explain why the force F is directly proportional to $\displaystyle pr^2v^2$
• Jul 2nd 2009, 06:42 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
When an object moves through a fluid , it experiences a retarding force due to turbulence . For a sphere of radius r moving with a speed of v in a fluid of density , p , the retarding force is given by

$\displaystyle F=kpr^2v^2$

By relating the retarding force to the transfer of momentum between the fluid and the sphere , explain why the force F is directly proportional to $\displaystyle pr^2v^2$

In a small time $\displaystyle \delta t$, the sphere moves a distance $\displaystyle \delta x$.We now assume that a volume of fluid proportional to the surface area of the sphere multiplied by the distance $\displaystyle \delta x$ has its velocity changed from zero to the velocity, $\displaystyle v$, of the sphere, due to impact with the sphere's surface. Since the SA of the sphere is proportional to the square of its radius, the mass of fluid affected in time $\displaystyle \delta t$ is therefore equal to

$\displaystyle k \rho r^2 \delta x$

where $\displaystyle \rho$ is the density of the fluid, and $\displaystyle k$ is a constant.

Its change in momentum is therefore

$\displaystyle k \rho r^2 \delta x v$

Therefore, using Newton's Second and Third Laws, the force that this mass of fluid exerts on the sphere is equal to the rate of change of its momentum; i.e.

$\displaystyle \frac{k \rho r^2 \delta x v}{\delta t}$

As $\displaystyle \delta t \rightarrow 0, \frac{\delta x}{\delta t}\rightarrow v$. So the force on the sphere is therefore

$\displaystyle k \rho r^2v^2$