In
mechanics we used to study the motion of the objects without considering the
resistive force they are present in out atmosphere. Any measurement we make in
real life will not be accurate if we don’t consider the resistive forces in the
Motion of an object.
Hence studying about the motion with resistive forces gives
us a clear understanding of what will really happen in nature. We will
encounter some mathematics to get the understanding.
It was experimentally understood that the resistive forces
are proportional to the quadratic relation with velocity.
First
to get a simple understanding we will make it a linear relation with velocity
in atmosphere.
Note:
All these Forces, velocities and acceleration are vector quantities. Dealing
with vector quantities is quite complicated than the scalars. So we always used
to split any motion into its components where we can deal it as the scalar
quantities.
So
we split a three dimensional motion into its fundamental components into two
horizontal components and a vertical component. That is in a Right Hand
Cartesian Coordinate system, we consider the x and y component as horizontal
components since it is coplanar with the ground. And we took the z component as
the vertical one as it is perpendicular to the plane of the ground. The ground
is considered to be flat because we are taking only a small portion of earth
that is a good approximation.
Thus
we are going to analyze the motion in three components but in two ways where
the motion in x and y direction is similar.
Part
– 1
Motion
of the object is coplanar with the plane of the ground. So, the velocity will
be considered as either the x or y component of velocity. Both components will
give similar results.
When
there is no other force in the horizontal direction using Newton’s Second Law,
m dv(x)/dt (or)
m dv(y)/dt = -m β v(x) or –m β v(y)
or
simply, we can write v(x) = v(y) = vx
Hence it gives us.
m
dvx /dt = - m β vx ,
β is constant that depends
on the shape and size of the object and on the resistive medium. The negative sign indicates that the resistive force
is in the opposite direction of the motion of the particle.
So, dvx /dt = -
β vx →
dvx / vx = - β
dt
Integrating dvx from “Vox” to “Vtx” and time from “0” to “t”
∫ dvx / vx = -β ∫ dt
→ Vtx =
Vox e^-βt ….. eq.(1)
From this result we can
know that,
·
If there is no
resistive force that is β = 0 then Vtx
= Vox .
The velocity will not change in that direction until there is an extra force
other than the resistive force.
·
The velocity
in the planar direction to the ground decreases exponentially with time.
Part – 2
Now we will talk about the
vertical motion where we will some non-intuitive concepts.
Motion
of the object is perpendicular to the plane of the ground. So, the velocity
will be considered as the z component of velocity that is perpendicular to the
ground.
In
the same way as we done before, Using Newton’s second law we can write the
equation of motion in perpendicular direction to the plane of ground as
follows,
but now we have to include
gravitational force. Taking velocity perpendicular to plane as vz
m dvz / dt =
- mg - β m vz
depending on the direction
of the velocity, the resistive force will change its direction.
dvz / dt = - g - β vz
→ dvz /( g + β vz ) = - dt
Integrating vz from “Voz” to “Vtz” and time from “0” to “t”
gives
→ Vtz = Voz
e^-βt +
g/β (e^-βt –
1) ……………. eq.(2)
Thus we have our second
equation.
From the equation (2) we
can understand,
·
Velocity in
perpendicular direction tends to a limiting value when time goes to infinite
value. The limiting value is equal to “-g/β”.
·
And this
limiting value is called the “The terminal velocity”.
·
From the
equation it is known that if the initial velocity is lower than the terminal
velocity, the object will tend to increase its speed until it reaches terminal
velocity.
·
But if the
initial velocity is higher than the terminal velocity the object will start to
slow down until it attends the terminal velocity.
·
The terminal
velocity of a skydiver in free fall is nearly equal to
200 kilometers per hour.
But to get more accurate
results, the resistive force should be treated as quadratic relations with
velocity. The mathematics to solve differential equation with quadratic
relations is too much complicated than the linear relations. However we will
give an attempt to that.
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