1. ## 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

$
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 $pr^2v^2$

2. Hello thereddevils
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

$
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 $pr^2v^2$
In a small time $\delta t$, the sphere moves a distance $\delta x$.We now assume that a volume of fluid proportional to the surface area of the sphere multiplied by the distance $\delta x$ has its velocity changed from zero to the velocity, $v$, 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 $\delta t$ is therefore equal to

$k \rho r^2 \delta x$

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

Its change in momentum is therefore

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

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.

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

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

$k \rho r^2v^2$