This is the idea of a body (having mass, "m") that moves with a constant speed and direction. The body moves steadily, irrespective of everything.
Two characteristics of a body in a defined space (say 0XZ) are its position, notated as P, and its velocity, V. Two properties of the body are its mass, m, and its momentum, mV. Newton used mathematics (calculus) to describe motions of bodies in space. He did this with this famous differential equation:
The above equation is an implicit description of all manners of motion of a body. Implicit means applicable to any mass that can be modelled as a body. Of all the manners of motion, uniform motion (for which momentum of the body, mV, is constant) is the least complex. With momentum constant, the above equation becomes two ideas expressed mathematically as:
We repeat that the equations apply to a body. Physically, both equations say the same thing. The equation above left states "the state of momentum of the body is unchanged as time proceeds." The other equation comments: " the sum of forces acting on the body equals zero, always." We are at the beginning, the very simplest level of mathematical physics. Let us make the equation (above left) specific to a case, and solve it.
Some physicists and engineers use the term, "trivial," to describe solutions of simple cases which they deem obvious, hence a waste of time to solve.. Below we solve, the most trivial case, the easiest application of Newton's Second Law of Motion. Newton described this simplest case with the words:
"A body at rest, remains at rest, provided (and as long as)
the sum of forces acting on it equals zero ."
The above statement has three prespectives of interpretation. The first is the physical reality of an "body at rest" on Earth; supported on a pedestal perhaps. The sketch to the right (without its surrounding notations) depicts that reality. The next perspective deals with time and event. The words imply that an observer, some person, noticed the body and that it is not moving, at some instant in time. Further implied is that the person continued to observe the zero sum of forces and continued state of rest.
The third prespective is that of mathematical physics. As foundation for the mathematical model, an inertial space ( 0XZ is enough), orthogonal coordinates and a vector basis must be defined. With those prerequisite tasks completed, the physics of the motion is stated mathematically as:
In the language of mathematics, we have a first order vector differential equation with the vector peoperty, momentum as its time-dependent variable and time as its independent variable. The domain of time starts at the initial instant of observation (t = 0+) and continues for all times thereafter. The equation states that the time derivative of momentum of the body always equals zero. The initial condition is: at t = 0+, the velocity of the body equals zero. To solve this differential equation means to determine Vbody(t) for all times t ≥ 0+.
It is obvious that a body at rest with no forces acting remains at rest, remains where it was originally. Some say to solve the obvious is a waste of time. Nonetheless we solve this case knowing what our solution must provide and along the way we will identify the steps which are the same trivial as well as complicated cases.
Solution is the mathematical task of integration of the differential equation. This procedure is said to be "classical" which means "at the very basis of mathemetics." The first order differential equation is very common. You will see this same procedure executed again and again in your studies. The steps even have names!
SEPARATE THE VARIABLES: As written above, the left side of the differential equation has two variables. The differential of momentum, d[mV(t)] is the dependent variable, time, t, is the independent variable. Upon multiplication of this simple equation by dt, the variables become separated:
INTEGRATE THE EQUATION: First apply integration symbols to both sides of the separated equation:
To specify the left-side limits, correspond them to the already stated, right-side limits. For the lower limit the body momentum, mV0+, (corresponding to time, t = 0+), is zero because the velocity is zero, mV. Since the upper limit of integration of time is indefinite; so also the left-side upper limit integration of momentum must be indefinite.
The limits now in place, we integrate.
INTERPRET RESULTS: The conclusion of our mathematics states:
Thus the velocity of the body was zero at t = 0+, and remained zero for all times thereafter.
We can take this a step further.
Above right we have another first order differential equation which requires an initial condition of position to be solved. Obviously the body was somewhere at t = 0+ . That place is P(t = 0+). So we repeat the process:
This last statement says for all times (including t = 0+) the position of the body is "where it was."
This first solution of Newton's differential equation is trivial in that the answer was known before we started. Nonetheless we solved its first order differential equation. Behaviors of very many physical systems can be expressed mathematically as first order differential equations. Solution steps are classical. So return... reread and confirm that these ideas of space, time frame, initial condition and such are correct. Commit this solution to memory; be able to reproduce it on a bar napkin to impress your date!
The next classical "motion of a body" solution is that of a body with constant momentum but non-zero initial velocity. Space and all are defined as before. The differential equation and initial condition are in the box.
Can you solve the differential equation? Give it a try! The solution for the body is:
P ( t ) = P0+ + V0+ t for all times, t > 0+.
It is likely that you have used this solution in physics but in a different form. The high school physics form is obtained by rearrangement of the above solution to form, P( t ) - P0+ = V0+t. Next ignore the specific idea of where the body "started" its motion... remove P0+ from the equation, to have: P( t ) = V0+t. Next to avoid vectors change P( t ) to S and V0+ to V, both scalars. That is pretty much the way the meaning and diversity of Newton's understanding of uniform motion is smoked and canned for use in high school physics: S = V t. Our next example, Yacht and Sea Buoy, illustrates (at a beginning level) uniform motion in which two-dimensional, relative motion is considered.
Yacht and Sea Buoy - A yacht sails from a dock Eastward with a speed of one mile per hour. Two miles northeast of the dock is a sea buoy. The event starts at t = 0+ with the yacht position and velocity, and the position of the sea buoy as indicated. Suppose a person on the yacht decided at the right time, he would dive ftom the yacht and swim to the sea buoy. Calculate the least distance of that swim and the time at which the swimmer should dive.
♦ A vector diagram is drawn with an origin placed at the dock, with 0XY coordinates and the unit vectors, I and J, are indicated. The yacht is shown a distance from the dock at the arbitrary time, t = t*.
It is a good habit to include time dependence by writing "( t )" with properties that change with time. The vector equation relating postions of the dock, yacht and buoy is:
Pyacht(t) + Psb/y(t) = Psb
This is a two-dimensional vector equation. When solved it will yield two pieces of information, that is two equations. The notation Psb/y is read, "position of the sea buoy (sb) with respect to the position of the yacht. (y). This is a vector that extends from the yacht to the sea buoy. To proceed, when writing an equation it is helpful to do the easy parts first. Of the three vectors we need, position of the buoy (strong>Pb) is the easiest to write, so we write it first.
Psb = 2 mi [cos(45°) I + sin(45°) J ]
The yacht, originally at the origin, moving Eastward at 1 mph for all times of our event; t > 0+.
Py(t) = [0 I + 0 J ] + [ 1mph t I ]
We put these results into the equation (with time in hours):
t I + Psb/y(t) = 2 mi [cos(45°) I + sin(45°) J ]
So everything is known except: Psb/y(t). We rearrange the equation:
Psb/y(t) = (√2 - t) I + √2 J
The distance between the yacht and buoy, D(t) is the length of the above vector:
D(t) = |Psb/y(t)| = [ (√2 - t)2 + (√2)2]1/2miles.
Complete the square to write D(t) as:
D(t) = [(t - √2)2 + 2]1/2miles.
So D(t) is least when t = tswim = √2 hours and the shortest distance is:
Dswim( tswim = √2 hours) = √2 miles.
Uniform motion is developed with a body as the system. An extension is to take a small amount of fluid as the system. Theat amount of water is but one piece of a larger, flowing stream of eawter, pieces besice pieces, beside othere. In this way the idea body is extended to that physical reality of a flowing fluid. The next example answers the question of velocity of oil in a pipeline. In addition the example introduces a mass equation.
Trans Alaska Pipeline - Some 800,000 barrels of oil per day are pumped 800 miles from Prudhome Bay to Valdez where it is loaded into tankers. At a location in the pipeline, a gage measures the average flow velocity of the oil passing through the four-foot diameter pipe. What is the average speed of the flowing oil?
♦ Let us imagine an event that starts right now, at the time, t = 0+, with us standing beside the flow gage. By this time tomorrow 800,000 barrels of oil will have passed. The pipe is four feet in diameter and there are 31.5gal/bbl.
At t = 0+, way upstream a distance L toward Prudhome, is a drop of oil moving in the pipe. That drop will be precisely here in 24 hours. The mass of oil contained by the pipe between here and the "drop" will have passed this location also. We know the volume of that mass, we use it and the pipe diameter mass to calculate the distance L.
This length is the magnitude of the position vector (at t = 0+) of the last drop of 800,000 barrels of oil pass this locaton by tomorrow. The actual pipe is not straight, but our answer will be correct. The initial and final position vectors of the drop are:
Uniform motion of a point in the oil is expressed as:
And upon substitution of our conditions,
At first it appears the answer is a negative speed while speed is zero or greater. The negative does not belong to the speed; it belongs to the direction ( - I ); meaning simply that the velocity is toward origin, the section of observation.
♦ A second way of determining the oil flow velocity is to visualize the filling of an empty oil barge. The image, the barge Douglas M., is empty and has the hold dimensions, 120 by 820 by 40 feet.
We take the mass of oil in the hold as our system. We write that mass symbolically at two times:
m( t* + Δt ) and at time, m( t = t*). Next substract the later from the former. That group is called a "difference." So divide the difference by the increment, Δt take the limit (as Δt → 0) and we have the derivative of mass which is the instantaneous rate of change of mass in the hold.
The rate of change of mass of oil within the tanker equals the constant mass rate (m-dotin) of oil that flows into the tank. Our simple differential equation becomes:
We developed this first order differential equation because some readers might not be familiar with it. The mass flow across the system boundary, m-dotin, equals the oil density time pipe area times flow speed. We now use constant hold geometry ( Abarge ) and constant density of oil to change the equation variable from mass to depth of oil, H(t).
The velocity of oil delivered in the pipe, Vin,avg is the answer we seek. As the tanker is filled, the depth of oil, H( t ), increases from zero to a second height. A simple calculation gives the oil depth after 800,000 barrels are loaded.
For this differential equation we have that the initial depth of oil in the tank, D( t = 0+) equals zero and the depth 24 hours later is D( t = 24hr) is 34.1 feet. So we separate variables and integrate:
Entering numbers...
The other aspect of zero velocity uniform motion is that the forces acting on the body sum to zero. The next example uses that fact, first with an Earth inertial reference then in a Moon inertial reference. See what you think?
Lunar "carry-off" Luggage -
By physical test on Earth it was determined a typical astronaut wearing a space suit could comfortably carry luggage having mass up to 30 kilograms. Astronauts headed to the moon don't load their luggage but they might have to unload it on the moon. So their bags can be packed a bit heavy. Earth gravity is 9.81 m/s2 and gravity on the moon is 1.62 m/s2.
Maximum Carry-Off: The lifting force an astonaut can apply on Earth can be applied equally on the moon. The force to support 30 kilograms on Earth is determined from Newton's Law of Acceleration applied vertical to Earth:
The astronaut can apply this force to unload on the moon. Hence the mass that might be unloaded is:
Notice the two events have separate inertial reference points. This is not a problem because no motion is considered, only force.
Both Metric and English Engineering Units are used in the United States. English Engineering Units have a peculiarity as is demonstrated below.
Ten Pounds of Potatoes -
The English Engineering Unit System states that force is a dimension. The magnitude of the unit force is defined to equal that force required to support a unit mass (one pound mass) at sea level. Thus to carry (in uniform motion) ten pounds mass of potatoes a person must exert a vertical force of ten pounds force. If we apply Newton's fact for uniform motion (0 = ΣF) in the vertical direction a useful identity among units (for potatoes and everything) results.
This equation seems peculiar because there is no unknown quantity. In actuality what the equation says is profound. It says that engineers who consider the dimensions mass, length, time and force as independent and then ascribe units arbitrarily will discover the dimensions are not independent and that their units are related but with a constant. In the English engineering system (F, m, L and t defined) the relationship among units is:
In contrast, the metric system defines only mass, length and time as dimensions. Units specified are the kilogram, meter and second. Force is left to be a "derived" entity. Were metric units used in this consideration we would obtain:
With length, the metric system is superior. The English Engineering System was pragmatic in locking the unit mass to be a unit force. It sure facilitates hydraulic engineering calculations.
Constant Momentum, Non-Zero Velocity Motion - Steps of the integration: Newton's First Law of Motion applies to bodies with constant momentum. His differential equation states that momentum (in 0XZ-space) is constant as long as the sum of forces equals zero ( ΣF = 0):
Newton's equation applies for times during which the body is observed. Specific epochs begin at the time, t = 0+, the instant observation commences. The above mathematical statement is inspecific; by itself it has no solution. To obtain a solution an "initial condition," the momentum of the body at the beginning (at t = 0+), is needed. Two cases for constant momentum of the mass are zero velocity and constant, non-zero velocity. The zero-velocity case was solved above. The solution for the second case proceeds as follows.