In lecture demonstrations, we do
measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.
The calculus of velocity-dependent frictional forces
When a body slides across a surface, the frictional force on it is approximately constant and given by
${\mu}_{\text{k}}N.$ Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by
${f}_{R}=\text{\u2212}bv,$
where
b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and
v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.
Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in
[link] . Newton’s second law in the vertical direction gives the differential equation
$mg-bv=m\frac{dv}{dt},$
where we have written the acceleration as
$dv\text{/}dt.$ As
v increases, the frictional force –
bv increases until it matches
mg . At this point, there is no acceleration and the velocity remains constant at the terminal velocity
${v}_{\text{T}}.$ From the previous equation,
$mg-b{v}_{\text{T}}=0,$
so
${v}_{\text{T}}=\frac{mg}{b}.$
We can find the object’s velocity by integrating the differential equation for
v . First, we rearrange terms in this equation to obtain
$\frac{dv}{g-(b\text{/}m)v}=dt.$
Assuming that
$v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields
where
$v\text{'}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\text{'}$ are dummy variables of integration. With the limits given, we find
Since
$\text{ln}A-\text{ln}B=\text{ln}(A\text{/}B),$ and
$\text{ln}(A\text{/}B)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain
A motorboat is moving across a lake at a speed
${v}_{0}$ when its motor suddenly freezes up and stops. The boat then slows down under the frictional force
${f}_{R}=\text{\u2212}bv.$ (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?
Solution
With the motor stopped, the only horizontal force on the boat is
${f}_{R}=\text{\u2212}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{\u2212}bv,$
which we can write as
$\frac{dv}{v}=-\frac{b}{m}dt.$
Integrating this equation between the time zero when the velocity is
${v}_{0}$ and the time
t when the velocity is
$v$ , we have
As time increases,
${e}^{\text{\u2212}bt\text{/}m}\to 0,$ and the position of the boat approaches a limiting value
${x}_{\text{max}}=\frac{m{v}_{0}}{b}.$
Although this tells us that the boat takes an infinite amount of time to reach
${x}_{\text{max}},$ the boat effectively stops after a reasonable time. For example, at
$t=10m\text{/}b,$ we have
Therefore, the boat’s velocity and position have essentially reached their final values.
With
${v}_{0}=4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and
$v=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}$ we have
$1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}){e}^{\text{\u2212}(b\text{/}m)(10\phantom{\rule{0.2em}{0ex}}\text{s})},$ so
Angular accelaration force,as the result of the rotatation of earth
Arzoodan
Due to the rotation of the earth the winds at the equator get deflected in opposite direction and therefore cause some currents in the Northern and Southern hemisphere which are opposite in there spin.
Varsha
Thats correct.
Arzoodan
if cos = x/y then sec = y/x
if sin = y/x then cosec = x/y
In a pulley system 2 boxes r hanging in both sides of pulley. An other box was joined to the left box through a rope & get accelerated downward. If all the boxes have same mass ,then what will be the acceleration of that system ?
the quantity of motion possessed by a body is called its momentum.by virtue of which a body can exert a force in the agency which tend to stop it .it is a common experience that stronger force is required to stop more massive body.also faster the body moves harder it is to stop it .this is why
Manoj
momentum is product of mass and velocity and it is denoted by "P".