4.3: Elementary Mechanics

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$trench-1995.jpg$
Contributed by William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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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

\[\label{eq:4.3.1} F=ma.\]

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 1 shows the other units. Consistent with Equation \ref{eq:4.3.1}, the unit of force in each system is defined to be the force required to impart an acceleration of (one unit of length)$/s^2$ to one unit of mass.

Table $\PageIndex{1}$
	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

\[\begin{array}{rl} g &=980\ \mbox{cm/s}^2 \hskip40pt \mbox{(cgs)} \\ g &=9.8\ \mbox{m/s}^2 \hskip48pt \mbox{(mks)} \\ g &=32\ \mbox{ft/s}^2 \hskip52pt \mbox{(British)}. \end{array}\nonumber \]

In general, the force $F$ in Equation \ref{eq:4.3.1} may depend upon $t$, $y$, and $y'$. Since $a=y''$, Equation \ref{eq:4.3.1} can be written in the form

\[\label{eq:4.3.2} my''=F(t,y,y'),\]

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 \ref{eq:4.3.2} can be recast as a first order equation. This is possible if $F$ does not depend on $y$, so Equation \ref{eq:4.3.2} is of the form

\[my''=F(t,y'). \nonumber \]

Letting $v=y'$ and $v'=y''$ yields a first order equation for $v$:

\[\label{eq:4.3.3} mv'=F(t,v).\]

Solving this equation yields $v$ as a function of $t$. If we know $y(t_0)$ for some time $t_0$, we can integrate $v$ to obtain $y$ as a function of $t$.

Equations of the form Equation \ref{eq:4.3.3} 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 $t\to\infty$. We call this limit the terminal velocity.

Example $\PageIndex{1}$

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 $v_0$.

Solution

The total force acting on the object is

\[\label{eq:4.3.4} F=-mg+F_1,\]

where $-mg$ is the force due to gravity and $F_1$ is the resisting force of the medium, which has magnitude $k|v|$, where $k$ is a positive constant. If the object is moving downward ($v\le 0$), the resisting force is upward (Figure $\PageIndex{1a}$), so

\[F_1=k|v|=k(-v)=-kv. \nonumber \]

On the other hand, if the object is moving upward ($v\ge 0$), the resisting force is downward (Figure $\PageIndex{1b}$), so

\[F_1=-k|v|=-kv.\nonumber\]

Thus, Equation \ref{eq:4.3.4} can be written as

\[\label{eq:4.3.5} F=-mg-kv,\]

regardless of the sign of the velocity.

From Newton’s second law of motion,

\[F=ma=mv',\nonumber\]

so Equation \ref{eq:4.3.5} yields

\[mv'=-mg-kv,\nonumber\]

\[\label{eq:4.3.6} v'+{k\over m}v=-g.\]

Since $e^{-kt/m}$ is a solution of the complementary equation, the solutions of Equation \ref{eq:4.3.6} are of the form $v=ue^{-kt/m}$, where $u'e^{-kt/m}=-g$, so $u'=-ge^{kt/m}$. Hence,

\[u=-{mg\over k} e^{kt/m}+c,\nonumber\]

\[\label{eq:4.3.7} v=ue^{-kt/m}=-{mg\over k}+ce^{-kt/m}.\]

Since $v(0)=v_0$,

\[v_0=-{mg\over k}+c,\nonumber\]

Figure $\PageIndex{1}$: Resistive forces

\[c=v_0+{mg\over k}\nonumber\]

and Equation \ref{eq:4.3.7} becomes

\[v=-{mg\over k}+\left(v_0+{mg\over k}\right) e^{-kt/m}.\nonumber\]

Letting $t\to\infty$ here shows that the terminal velocity is

\[\lim_{t\to\infty} v(t)=-{mg\over k},\nonumber\]

which is independent of the initial velocity $v_0$ (Figure $\PageIndex{2}$).

Example $\PageIndex{2}$: 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

\[30v'=-960-3v,\nonumber\]

which we rewrite as

\[v'+{1\over 10}v=-32.\nonumber\]

Solutions of Mv' = -mg - kv — Figure $\PageIndex{2}$: Solutions of $mv'=-mg-kv$

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 $u'=-32e^{t/10}$. Hence,

\[u=-320 e^{t/10}+c,\nonumber\]

\[\label{eq:4.3.8} v=ue^{-t/10}=-320+ce^{-t/10}.\]

The initial velocity is 60 ft/s in the upward (positive) direction; hence, $v_0=60$. Substituting $t=0$ and $v=60$ in Equation \ref{eq:4.3.8} yields

\[60=-320+c,\nonumber\]

so $c=380$, and Equation \ref{eq:4.3.8} becomes

\[v=-320+380e^{-t/10}\ \mbox{ft/s}\nonumber\]

The terminal velocity is

\[\lim_{t\to\infty}v(t)=-320\mbox{ ft/s.}\nonumber\]

Example $\PageIndex{3}$

A 10 kg mass is given an initial velocity $v_0\le0$ 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,

\[\label{eq:4.3.9} mv'=-mg+kv^2,\]

where $k$ is a constant. Since the magnitude of the resistance is 8 N when $v=2$ m/s,

\[k(2^2)=8, \nonumber \]

so $k=2\ \mbox{N-s}^2/\mbox{m}^2$. Since $m=10$ and $g=9.8$, Equation \ref{eq:4.3.9} becomes

\[\label{eq:4.3.10} 10v'=-98+2v^2=2(v^2-49).\]

If $v_0=-7$, then $v\equiv-7$ for all $t\ge0$. If $v_0\ne-7$, we separate variables to obtain

\[\label{eq:4.3.11} {1\over v^2-49}v'={1\over5},\]

which is convenient for the required partial fraction expansion

\[\label{eq:4.3.12} \frac{1}{v^2-49} =\frac{1}{(v-7)(v+7)} ={1\over 14}\left[{1\over v-7} -{1\over v+7}\right].\]

Substituting Equation \ref{eq:4.3.12} into Equation \ref{eq:4.3.11} yields

\[{1\over14}\left[{1\over v-7}-{1\over v+7}\right]v'={1\over5},\nonumber\]

\[\left[{1\over v-7}-{1\over v+7}\right]v'={14\over5}.\nonumber\]

Integrating this yields

\[\ln |v-7|-\ln|v+7|=14t/5+k.\nonumber\]

Therefore

\[\left|{v-7\over v+7}\right|=e^ke^{14t/5}.\nonumber\]

Since Theorem [thmtype:2.3.1} implies that $(v-7)/(v+7)$ cannot change sign (why?), we can rewrite the last equation as

\[\label{eq:4.3.13} {v-7\over v+7}=ce^{14t/5},\]

which is an implicit solution of Equation \ref{eq:4.3.10}. Solving this for $v$ yields

\[\label{eq:4.3.14} v=-7{c+e^{-14t/5}\over c-e^{-14t/5}}.\]

Since $v(0)=v_0$, it Equation \ref{eq:4.3.13} implies that

\[c={v_0-7\over v_0+7}.\nonumber\]

Substituting this into Equation \ref{eq:4.3.14} and simplifying yields

\[v=-7{v_0(1+e^{-14t/5}-7(1-e^{-14t/5}\over v_0(1-e^{-14t/5}-7(1+e^{-14t/5}}.\nonumber\]

Since $v_0\le0$, $v$ is defined and negative for all $t>0$. The terminal velocity is

\[\lim_{t\to\infty} v(t)=-7\ \mbox{m/s},\nonumber\]

independent of $v_0$. More generally, it can be shown (Exercise 4.3.11) that if $v$ is any solution of Equation \ref{eq:4.3.9} such that $v_0\le0$ then

\[\lim_{t\to\infty}v(t)=-\sqrt{mg\over k}\nonumber\]

(Figure $\PageIndex{3})$.

Solutions of mv'=-mg+kv^2,\ v(0)=v_0\le0 — Figure $\PageIndex{3}$: Solutions of $mv'=-mg+kv^2,\ v(0)=v_0\le0$

Example $\PageIndex{4}$:

A 10-kg mass is launched vertically upward from Earth’s surface with an initial velocity of $v_0$ 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 $0\leq t<T$ and $v(T)=0$; therefore, we replace Equation \ref{eq:4.3.10} by

\[\label{eq:4.3.15} 10v'=-98-2v^2.\]

Separating variables yields

\[{5\over v^2+49}v'=-1,\nonumber\]

and integrating this yields

\[{5\over7}\tan^{-1}{v\over7}=-t+c.\nonumber\]

(Recall that $\tan^{-1}u$ is the number $\theta$ such that $-\pi/2 < \theta < \pi/2$ and $\tan \theta=u$.) Since $v(0)=v_0$,

\[c={5\over7}\tan^{-1}{v_0\over7},\nonumber\]

so $v$ is defined implicitly by

\[\label{eq:4.3.16} {5\over7} \tan^{-1}{v\over7}=-t+{5\over7} \tan^{-1}{v_0\over7}, \quad 0\le t\le T.\]

Figure $\PageIndex{4}$: Solutions of $\PageIndex{15}$ for various $v_{0}>0$

Solving this for $v$ yields

\[\label{eq:4.3.17} v=7\tan\left(-{7t\over5}+\tan^{-1}{v_0\over7}\right).\]

Using the identity

\[\tan(A-B)={\tan A-\tan B\over1+\tan A\tan B}\nonumber\]

with $A=\tan^{-1}(v_0/7)$ and $B=7t/5$, and noting that $\tan(\tan^{-1}\theta)=\theta$, we can simplify Equation \ref{eq:4.3.17} to

\[v=7{v_0-7\tan(7t/5)\over7+v_0\tan(7t/5)}.\nonumber\]

Since $v(T)=0$ and $\tan^{-1}(0)=0$, Equation \ref{eq:4.3.16} implies that

\[-T+{5\over7} \tan^{-1}{v_0\over7}=0.\nonumber\]

Therefore

\[T={5\over7} \tan^{-1}{v_0\over7}.\nonumber\]

Since $\tan^{-1}(v_0/7)<\pi/2$ for all $v_0$, the time required for the mass to reach its maximum altitude is less than

\[{5\pi\over 14} \approx 1.122\ \mbox{s}\nonumber\]

regardless of the initial velocity. Figure $\PageIndex{4}$ shows graphs of $v$ over $[0,T]$ for various values of $v_0$.

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

\[\label{eq:4.3.18} F=-{K\over(y+R)^2},\]

where $R$ is Earth’s radius (Figure $\PageIndex{5}$).

Since $F=-mg$ when $y=0$, setting $y=0$ in Equation \ref{eq:4.3.18} yields

\[-mg=-{K\over R^2};\nonumber\]

therefore $K=mgR^2$ and Equation \ref{eq:4.3.18} can be written more specifically as

\[\label{eq:4.3.19} F=-{mgR^2\over(y+R)^2}.\]

From Newton’s second law of motion,

\[F=m{d^2y\over dt^2},\nonumber\]

so Equation \ref{eq:4.3.19} implies that

\[\label{eq:4.3.20} {d^2y\over dt^2}=-{gR^2\over(y+R)^2}.\]

We’ll show that there’s a number $v_e$, called the escape velocity, with these properties:

If $v_0\ge v_e$ 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 $\lim_{t\to\infty}y(t)=\infty$ in this case? For a proof, see Exercise 4.3.20.)
If $v_0 < v_e$ then $v(t)$ decreases to zero and becomes negative. Therefore the vehicle attains a maximum altitude $y_{m}$ and falls back to Earth.

Since Equation \ref{eq:4.3.20} 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

\[{d^2y\over dt^2}={dv\over dt}={dv\over dy}{dy\over dt}=v{dv\over dy}. \nonumber \]

Substituting this into Equation \ref{eq:4.3.20} yields the first order separable equation

\[\label{eq:4.3.21} v{dv\over dy}=-{gR^2\over(y+R)^2}.\]

When $t=0$, the velocity is $v_0$ and the altitude is $h$. Therefore we can obtain $v$ as a function of $y$ by solving the initial value problem

\[v{dv\over dy}=-{gR^2\over(y+R)^2},\quad v(h)=v_0.\nonumber\]

Integrating Equation \ref{eq:4.3.21} with respect to $y$ yields

\[\label{eq:4.3.22} {v^2\over 2}={gR^2\over y+R}+c.\]

Since $v(h)=v_0$,

\[c={v_0^2\over 2}-{gR^2\over h+R},\nonumber\]

so Equation \ref{eq:4.3.22} becomes

\[\label{eq:4.3.23} {v^2\over 2}={gR^2\over y+R}+\left({v_0^2\over 2}- {gR^2\over h+R}\right).\]

\[v_0 \ge\left({2gR^2\over h+R}\right)^{1/2},\nonumber\]

the parenthetical expression in Equation \ref{eq:4.3.23} is nonnegative, so $v(y)>0$ for $y>h$. This proves that there’s an escape velocity $v_e$. We’ll now prove that

\[v_e=\left({2gR^2\over h+R}\right)^{1/2}\nonumber\]

by showing that the vehicle falls back to Earth if

\[\label{eq:4.3.24} v_0 <\left({2gR^2\over h+R}\right)^{1/2}.\]

If Equation \ref{eq:4.3.24} holds then the parenthetical expression in Equation \ref{eq:4.3.23} is negative and the vehicle will attain a maximum altitude $y_m>h$ that satisfies the equation

\[0={gR^2\over y_m+R}+\left({v_0^2\over 2}- {gR^2\over h+R}\right).\nonumber\]

The velocity will be zero at the maximum altitude, and the object will then fall to Earth under the influence of gravity.