Download Physics notes Newton's law and more topics and more Cheat Sheet Physics in PDF only on Docsity!
Newton's second law in action
body is known. In many cases, the nature of the force acting on a
It might depend on time,
In such cases, Newton's law becomes anbut its dependence is known from experiment.position, velocity, or some combination of these,
equation
of motion
which we can solve.
The solution
and velocity.any time, as long as we know its initial positionallows us to predict the position of the body at
This predictive quality is the main
in this way.In this section, we will use Newton’s second lawpower of Newton’s law.
We will consider the case of one-
with and without damping.dimensional motion under a constant force, both
This has many
applications, one of which is vertical motion
near earth’s surface.
Motion under a constant force
not depend on the body's velocity.is the same no matter where the body is, and doesconstant, we mean that it does not vary over time,When we say that the force acting on a body is
In other
words,
F
is just a number. The following graph
shows a case in which the constant force
happens
to be in the positive direction:
F
x the graph of the position of the body versus time, We are going to use Newton's law to show that ( t ), is a parabola.
We begin with Newton’s law,
F=ma
. Using the
derivative ofdefinition of the acceleration as the second
x ( t ), we find the following equation
of motion:
m
d
(^) 2 x^ ( t )
dt
2
F
This is a
differential equation
for the function
x ( t ). That is, it is an equation whose solution is a
whole
function of time
, not just an algebraic
number.
It says that
x
( t ) is a function whose
constant,second derivative with respect to time is a
F/m
. Therefore, we know what its first
and findderivative must be; that is, we can integrate once
dx
t )
dt
m F
t
constant
Now, the left-hand side is the velocity.where the constant does not depend on time.
So the
velocity at some particular time.constant must be fixed by the value of the
For simplicity,
let's suppose that this time is
t =0, and that the
velocity at that time is
v
0 .
Substituting
t =0 into
is both sides, we find that the value of the constant
v 0
. The velocity is therefore given by
v ( t )
m F
t
v
0
.
We then integrate again, and find
x ( t ) (^) =
m F
t 2
v 0 t
another constant
again take to beposition at another particular time, which we mayThe new constant is fixed by the value of the
t
If the position then is
x 0,
then the value of the new constant is
x
0 .
The
is thus given byfinal answer for the position as a function of time
x ( t ) (^) =
F
m
t 2
v 0 (^) t
(^) x
0
time, as long as we know the values of the initialThis formula allows us to find the position at any
position and velocity.
The reason we have to
know
two
quantities is because Newton's law
gives rise to a
second-order
differential equation.
Suppose the body is initially at rest atSpecial case: zero initial velocity, positive forcewill do this by considering some special cases.Let's try to picture such a motion physically. Wethe second derivative.That is, the highest derivative which appears is
x 0 =0, and
the force acts in the positive
x
direction.
Then
speed up in the direction of the force.we know what will happen - the body will just
The
position as a function of time is given by
x ( t )
F
m
t 2
.
This is (one half of) a parabola pointing upwards:
t
x
x(t)
executing the input.easily change the values of the parameters beforedocument and paste them into MAPLE.) You can
If you wish, you can also
plot
the
velocity
by
changing
s o l n
to
diff(soln,t)
in the above line.
near the surface of the earth,the relevant fact:interaction until later, you may be familiar withAlthough we will not study the gravitationalmoving vertically near the surface of the earth.above applies directly is the case of a body A very important physical situation to which the Application: vertical motion near earth
all
bodies (whatever
approximately 9.81 m stheir inertial mass) have an acceleration of
in the downward
are negligible.This is true as long as the effects of air resistancedirection.
This value is given a special
symbol,
g
, and is called the
gravitational acceleration at earth’s surface:
g
9.81 m s
due to gravity ispositive values pointing upwards. Then the forceLet's orient our coordinates vertically, with
F
mg
downwards. The minus sign indicates that the force is directed
(If we had oriented the coordinates
forminus sign would not be present in the equationwith positive values pointing downwards, the
F
-3^ -2^ -1 0 1
f
the above formulas directly, withposition, or velocity. We can therefore take over This force is constant - does not depend on time,
mg
substituted
for
F
. In particular, we find that
- a body fired upwards with velocity
v
0
reaches a
maximum height of
v
2
0
g
above its starting height, at time
v
0^
g
The path in the
t-x
plane is given by the parabola
that the pathshown two figures ago. It is important to realize
in space
is a straight line, not a
parabola.
We are considering vertical motion
only
, at present.
(Later, when
we
come
to
pathwill find that a projectile can move on a parabolicconsider motion in more than one dimension, we
in space
That parabolic path is
not the
same
as the present one in the
t-x
plane - don't get
position,showing this vertical motion along with graphs ofThe resources for this section contain a moviethem confused!)
velocity
and
acceleration.
For
same quantities for a body dropped from rest.comparison, there is also a movie showing the
Resistive forces
along a track with friction present.to motion. For example, a body could be slidingIn many physical cases, there is some resistance
Or, a body
air resistance.could be moving vertically near the earth, with
proportional to theIn many cases, the force resisting the motion is
velocity
of
the
body.
Mathematically, this is written
F
res
(^) bv
The quantity
b
is a positive constant, whose value
depends
on
the
properties
of
the
material
providing the resistance.
It is not a fundamental
opposite the force resists the motion, so is directedconstant of nature. The minus sign indicates that
to the velocity.
the body is the sum of the constant forceinvolved. We will suppose that the total force onsolving Newton’s law when such a force isWe would like to illustrate the procedure of
F
that
forcewe have just been considering, and the resistive
F
res
F
net
F
F
res
called theThe body will then move with a constant velocitybalance the resistive force, giving zero net force.goes on, the external constant force will justis easy to see one aspect of the solution. As timebeginning the mathematics. In the present case, itto get an idea of the nature of the solution, beforeIt is always a good idea to use physical intuition
terminal velocity
v
t .
x ( t ).
We integrate our expression for
v
( t ) once,
obtaining
x(t)
b F
t (^) −
b m
ã ååå ä v^ 0 −
b F
í ììì ë
exp
ã åååä −^
m b
t í ì^ ììë
constant.
Let’s say that
x=x
0
at
t =0.
Substituting into the
above expression yields
constant
x
0
b m
ã ååå ä v^ 0 −
b F
í ììì ë
.
Hence, the full solution for
x ( t ) is
x ( t )
x 0
b F
t
b m
ã ååå ä v^ 0 −
b F
í ììì ë ã åååä 1
exp
ã åååä −^
m b
t í ì^ ìì ë í ìììë
To answer our question, we substitute
t max
into
our expression for
x
( t ), obtaining
x
max^
x 0
mv
0
b
Fm b
(^2)
ln
ã ååååä 1
bv
0
F
í ìììì ë
.^
Here is a plot of
x(t)
showing the effect of
noticeably from the ideal parabolic form.We see that the curve with resistance deviatesbut no resistance is shown in red for comparison.resistance; a plot with the same initial conditions
slope =
t
x(t)
no resistance
with resistance
v 0
velocity by changingexecuting. If you wish, you can also plot theeasily change the values of the parameters beforedocument and paste them into MAPLE.) You can (Here’s how to copy these lines from the presentsoln),t=0..2); plot(subs({m=1,F=-1,b=1,x[0]=0,v[0]=1}, The next line plots a solution:x(t)));bdiff(x(t),t),x(0)=x[0],D(x)(0)=v[0]}, soln:=op(2,dsolve({mdiff(x(t),t$2)=F-present differential equation symbolically: Here is a MAPLE input line which solves the
soln
to
diff(soln,t)
in
ways to check the answers we obtained in this In the next section, we will consider a couple of Checking our answers: the above line.
section.
In particular, we will see how to make
sure that the units are correct (
dimensional
analysis
), and also how to make sure that the
the damping to zero in the more general case.case of zero damping is recovered when we set
