4.3: Elementary Mechanics
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Newton's Second Law of Motion
In this section we consider an object with constant mass m moving along a line under a force F. Let y=y(t) be the displacement of the object from a reference point on the line at time t, and let v=v(t) and a=a(t) be the velocity and acceleration of the object at time t. Thus, v=y′ and a=v′=y″, where the prime denotes differentiation with respect to t. Newton’s second law of motion asserts that the force F and the acceleration a are related by the equation
F=ma.
Note: Units
In applications there are three main sets of units in use for length, mass, force, and time: the cgs, mks, and British systems. All three use the second as the unit of time. Table 4.3.1 shows the other units. Consistent with Equation ???, the unit of force in each system is defined to be the force required to impart an acceleration of (one unit of length)/s2 to one unit of mass.
Set | Length | Force | Mass |
---|---|---|---|
cgs | centimeter (cm) | dyne (d) | gram (g) |
mks | meter (m) | newton (N) | kilogram (kg) |
British | foot (ft) | pound (lb) | slug (sl) |
If we assume that Earth is a perfect sphere with constant mass density, Newton’s law of gravitation (discussed later in this section) asserts that the force exerted on an object by Earth’s gravitational field is proportional to the mass of the object and inversely proportional to the square of its distance from the center of Earth. However, if the object remains sufficiently close to Earth’s surface, we may assume that the gravitational force is constant and equal to its value at the surface. The magnitude of this force is mg, where g is called the acceleration due to gravity. (To be completely accurate, g should be called the magnitude of the acceleration due to gravity at Earth’s surface.) This quantity has been determined experimentally. Approximate values of g are
g=980 cm/s2(cgs)g=9.8 m/s2(mks)g=32 ft/s2(British).
In general, the force F in Equation ??? may depend upon t, y, and y′. Since a=y″, Equation ??? can be written in the form
my″=F(t,y,y′),
which is a second order equation. We’ll consider this equation with restrictions on F later; however, since Chapter 2 dealt only with first order equations, we consider here only problems in which Equation ??? can be recast as a first order equation. This is possible if F does not depend on y, so Equation ??? is of the form
my″=F(t,y′).
Letting v=y′ and v′=y″ yields a first order equation for v:
mv′=F(t,v).
Solving this equation yields v as a function of t. If we know y(t0) for some time t0, we can integrate v to obtain y as a function of t.
Equations of the form Equation ??? occur in problems involving motion through a resisting medium.
Motion Through a Resisting Medium Under Constant Gravitational Force
Now we consider an object moving vertically in some medium. We assume that the only forces acting on the object are gravity and resistance from the medium. We also assume that the motion takes place close to Earth’s surface and take the upward direction to be positive, so the gravitational force can be assumed to have the constant value −mg. We’ll see that, under reasonable assumptions on the resisting force, the velocity approaches a limit as t→∞. We call this limit the terminal velocity.
Example 4.3.1
An object with mass m moves under constant gravitational force through a medium that exerts a resistance with magnitude proportional to the speed of the object. (Recall that the speed of an object is |v|, the absolute value of its velocity v.) Find the velocity of the object as a function of t, and find the terminal velocity. Assume that the initial velocity is v0.
Solution
The total force acting on the object is
F=−mg+F1,
where −mg is the force due to gravity and F1 is the resisting force of the medium, which has magnitude k|v|, where k is a positive constant. If the object is moving downward (v≤0), the resisting force is upward (Figure 4.3.1a ), so
F1=k|v|=k(−v)=−kv.
On the other hand, if the object is moving upward (v≥0), the resisting force is downward (Figure 4.3.1b , so
F1=−k|v|=−kv.
Thus, Equation ??? can be written as
F=−mg−kv,
regardless of the sign of the velocity.
From Newton’s second law of motion,
F=ma=mv′,
so Equation ??? yields
mv′=−mg−kv,
or
v′+kmv=−g.
Since e−kt/m is a solution of the complementary equation, the solutions of Equation ??? are of the form v=ue−kt/m, where u′e−kt/m=−g, so u′=−gekt/m. Hence,
u=−mgkekt/m+c,
so
v=ue−kt/m=−mgk+ce−kt/m.
Since v(0)=v0,v0=−mgk+c,so
c=v0+mgk
and Equation ??? becomes
v=−mgk+(v0+mgk)e−kt/m.
Letting t→∞ here shows that the terminal velocity is
limt→∞v(t)=−mgk,
which is independent of the initial velocity v0 (Figure 4.3.2 ).
Example 4.3.2 : Terminal Velocity
A 960-lb object is given an initial upward velocity of 60 ft/s near the surface of Earth. The atmosphere resists the motion with a force of 3 lb for each ft/s of speed. Assuming that the only other force acting on the object is constant gravity, find its velocity v as a function of t, and find its terminal velocity.
Solution
Since mg=960 and g=32, m=960/32=30. The atmospheric resistance is −3v lb if v is expressed in feet per second. Therefore
30v′=−960−3v,
which we rewrite asv′+110v=−32.
Since e−t/10 is a solution of the complementary equation, the solutions of this equation are of the form v=ue−t/10, where u′e−t/10=−32, so u′=−32et/10. Hence,
u=−320et/10+c,
so
v=ue−t/10=−320+ce−t/10.
The initial velocity is 60 ft/s in the upward (positive) direction; hence, v0=60. Substituting t=0 and v=60 in Equation ??? yields
60=−320+c,
so c=380, and Equation ??? becomes
v=−320+380e−t/10 ft/s
The terminal velocity is
limt→∞v(t)=−320 ft/s.
Example 4.3.3
A 10 kg mass is given an initial velocity v0≤0 near Earth’s surface. The only forces acting on it are gravity and atmospheric resistance proportional to the square of the speed. Assuming that the resistance is 8 N if the speed is 2 m/s, find the velocity of the object as a function of t, and find the terminal velocity.
Solution
Since the object is falling, the resistance is in the upward (positive) direction. Hence,
mv′=−mg+kv2,
where k is a constant. Since the magnitude of the resistance is 8 N when v=2 m/s,
k(22)=8,
so k=2 N-s2/m2. Since m=10 and g=9.8, Equation ??? becomes
10v′=−98+2v2=2(v2−49).
If v0=−7, then v≡−7 for all t≥0. If v0≠−7, we separate variables to obtain
1v2−49v′=15,
which is convenient for the required partial fraction expansion
1v2−49=1(v−7)(v+7)=114[1v−7−1v+7].
Substituting Equation ??? into Equation ??? yields
114[1v−7−1v+7]v′=15,
so
[1v−7−1v+7]v′=145.
Integrating this yields
ln|v−7|−ln|v+7|=14t/5+k.
Therefore
|v−7v+7|=eke14t/5.
Since Theorem 2.3.1 implies that (v−7)/(v+7) cannot change sign (why?), we can rewrite the last equation as
v−7v+7=ce14t/5,
which is an implicit solution of Equation ???. Solving this for v yields
v=−7c+e−14t/5c−e−14t/5.
Since v(0)=v0, it Equation ??? implies that
c=v0−7v0+7.
Substituting this into Equation ??? and simplifying yields
v=−7v0(1+e−14t/5)−7(1−e−14t/5)v0(1−e−14t/5)−7(1+e−14t/5).
Since v0≤0, v is defined and negative for all t>0. The terminal velocity is
limt→∞v(t)=−7 m/s,
independent of v0. More generally, it can be shown (Exercise 4.3.11) that if v is any solution of Equation ??? such that v0≤0 then
limt→∞v(t)=−√mgk.
This is demonstrated in Figure 4.3.3 .
Example 4.3.4
A 10-kg mass is launched vertically upward from Earth’s surface with an initial velocity of v0 m/s. The only forces acting on the mass are gravity and atmospheric resistance proportional to the square of the speed. Assuming that the atmospheric resistance is 8 N if the speed is 2 m/s, find the time T required for the mass to reach maximum altitude.
Solution
The mass will climb while v>0 and reach its maximum altitude when v=0. Therefore v>0 for 0≤t<T and v(T)=0; therefore, we replace Equation ??? by
10v′=−98−2v2.
Separating variables yields
5v2+49v′=−1,
and integrating this yields
57tan−1v7=−t+c.
(Recall that tan−1u is the number θ such that −π/2<θ<π/2 and tanθ=u.) Since v(0)=v0,
c=57tan−1v07,
so v is defined implicitly by57tan−1v7=−t+57tan−1v07,0≤t≤T.
Solving this for v yields
v=7tan(−7t5+tan−1v07).
Using the identity
tan(A−B)=tanA−tanB1+tanAtanB
with A=tan−1(v0/7) and B=7t/5, and noting that tan(tan−1θ)=θ, we can simplify Equation ??? to
v=7v0−7tan(7t/5)7+v0tan(7t/5).
Since v(T)=0 and tan−1(0)=0, Equation ??? implies that
−T+57tan−1v07=0.
Therefore
T=57tan−1v07.
Since tan−1(v0/7)<π/2 for all v0, the time required for the mass to reach its maximum altitude is less than
5π14≈1.122 s
regardless of the initial velocity. Figure 4.3.4 shows graphs of v over [0,T] for various values of v0.
Escape Velocity
Suppose a space vehicle is launched vertically and its fuel is exhausted when the vehicle reaches an altitude h above Earth, where h is sufficiently large so that resistance due to Earth’s atmosphere can be neglected. Let t=0 be the time when burnout occurs. Assuming that the gravitational forces of all other celestial bodies can be neglected, the motion of the vehicle for t>0 is that of an object with constant mass m under the influence of Earth’s gravitational force, which we now assume to vary inversely with the square of the distance from Earth’s center; thus, if we take the upward direction to be positive then gravitational force on the vehicle at an altitude y above Earth is
F=−K(y+R)2,
where R is Earth’s radius (Figure 4.3.5 ).
Since F=−mg when y=0, setting y=0 in Equation ??? yields
−mg=−KR2;
therefore K=mgR2 and Equation ??? can be written more specifically as
F=−mgR2(y+R)2.
From Newton’s second law of motion,
F=md2ydt2,
so Equation ??? implies that
d2ydt2=−gR2(y+R)2.
We’ll show that there’s a number ve, called the escape velocity, with these properties:
- If v0≥ve then v(t)>0 for all t>0, and the vehicle continues to climb for all t>0; that is, it “escapes” Earth. (Is it really so obvious that limt→∞y(t)=∞ in this case? For a proof, see Exercise 4.3.20.)
- If v0<ve then v(t) decreases to zero and becomes negative. Therefore the vehicle attains a maximum altitude ym and falls back to Earth.
Since Equation ??? is second order, we cannot solve it by methods discussed so far. However, we are concerned with v rather than y, and v is easier to find. Since v=y′ the chain rule implies that
d2ydt2=dvdt=dvdydydt=vdvdy.
Substituting this into Equation ??? yields the first order separable equation
vdvdy=−gR2(y+R)2.
When t=0, the velocity is v0 and the altitude is h. Therefore we can obtain v as a function of y by solving the initial value problem
vdvdy=−gR2(y+R)2,v(h)=v0.
Integrating Equation ??? with respect to y yields
v22=gR2y+R+c.
Since v(h)=v0,
c=v202−gR2h+R,
so Equation ??? becomes
v22=gR2y+R+(v202−gR2h+R).
If
v0≥(2gR2h+R)1/2,
the parenthetical expression in Equation ??? is nonnegative, so v(y)>0 for y>h. This proves that there’s an escape velocity ve. We’ll now prove that
ve=(2gR2h+R)1/2
by showing that the vehicle falls back to Earth if
v0<(2gR2h+R)1/2.
If Equation ??? holds then the parenthetical expression in Equation ??? is negative and the vehicle will attain a maximum altitude ym>h that satisfies the equation
0=gR2ym+R+(v202−gR2h+R).
The velocity will be zero at the maximum altitude, and the object will then fall to Earth under the influence of gravity.