Download Dynamics

Dynamics
What You Will Learn


How to relate position, velocity, acceleration and time.
How to relate force and motion.
Introduction
The study of dynamics is the study of motion, including the speed and acceleration of a part, the path it
takes, and the orientation of the part during its journey. Position, velocity, acceleration and time can be
related through the techniques of calculus, and will be the focus of the first part of this section. The
second half of the section deals with how to relate force to motion. There are several methods available
to engineers, and none will be discussed in detail. Instead, simple problems will be solved by each
method so that we can compare and contrast the benefits of using each so that you will be able to pick
the best method for a problem.
A typical industrial application might be setting requirements for motors for a robot based on the type of
work it will do, and the speed at which the task must be accomplished. Another area where dynamics
concepts play a large role is in the design of space missions, whether a simple orbit, or the meeting up of
a shuttle with the space station, the trajectories must be carefully choreographed using the tools
introduced in this section.
Relating Position, Velocity, Acceleration and Time
Often times, position as a function of time may be known, but not acceleration or velocity. In other
situations acceleration may be known as a function of velocity, but it is desired to know the position over
time. The following examples may be useful in demonstrating that this is not just a theoretical problem.
Examples
1. Braking Distance
a. Given: A top FSAE student competition car going 60kph can decelerate at about 1g.
b. Find: The distance needed to stop.
c. Mathematical rephrasing of question: Given a constant acceleration, what is the change
in position, x if velocity changes from 60kpm to zero? This will require knowing velocity
as a function of position, v(x).
2. Mars Landing
a. Given: A scientific mission to Mars will be out of communication during descent.
b. Find: The time to touchdown, so that the mission can be planned accordingly.
c. Mathematical rephrasing of question: The gravitational force will increase as the lander
approaches the surface, so the acceleration will be a function of position, a(x). We want
to know v(x), where x is the “height” of the surface.
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3. Human Cannonball
a. Given: You are pursuing a career as a human cannonball. You know the launch velocity,
and can assume gravity is a constant.
b. Find: how far away to put landing pad.
c. Mathematical rephrasing of question: You have vertical acceleration as a function of time
a(t), but you need to get distance as a function of time x(t).
How can we solve these problems? We must start with the most basic of definitions, first that velocity is
the change in distance over time:
change in distance
velocity =
change in time
In the language of calculus, we can write this as
𝒅𝒙 𝒕
𝒗 𝒕 =
𝒅𝒕
Similarly, acceleration can be defined as the change in velocity over the change in time
change in velocity
acceleration =
change in time
In the language of calculus, we can write this as
𝒗 𝒕
𝒂 𝒕 =
𝒅𝒕
We can combine these equations to get acceleration in terms of position:
𝑣 𝑡
𝑑 𝑑𝑥 𝑡
𝑑2 𝑥 𝑡
𝑎 𝑡 =
=
=
𝑑𝑡
𝑑𝑡 𝑑𝑡
𝑑𝑡 2
In order to reverse these calculations, we integrate, which is the opposite of differentiation:
𝑑𝑥 𝑡
𝑣 𝑡 =
→ ∫ 𝑑𝑥 = ∫ 𝑣(𝑡)𝑑𝑡 → 𝒙 𝒕 = ∫ 𝒗 𝒕 𝒅𝒕
𝑑𝑡
Similarly, we can find velocity by integrating acceleration:
𝑑𝑣 𝑡
𝑎 𝑡 =
→ ∫ 𝑑𝑣 = ∫ 𝑎(𝑡)𝑑𝑡 → 𝒗 𝒕 = ∫ 𝒂 𝒕 𝒅𝒕
𝑑𝑡
There is one more relationship that we need in order to be able to work with any combination of
dependent and independent variables. Start with the definition of acceleration
𝑑𝑣
𝑎=
𝑑𝑡
Now we will multiply by a clever form of one.
𝑑𝑣 𝑑𝑥
𝑎=
⋅
𝑑𝑡 𝑑𝑥
1
Now rearrange the numerator on the right hand side.
𝑑𝑥 𝑑𝑣
𝑑𝑣
𝑎=
⋅
=𝑣
𝑑𝑡 𝑑𝑥
𝑑𝑥
𝑣
This relationship allows us to relate acceleration to velocity when velocity is a function of position.
The chart below is a graphical representation of the relationship between position velocity acceleration
and time. The arrows in front of each equation indicate which way the equation works
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→òadx=òvdv
←a = vdv/dx
a(x)
→òdx=ò(v/a)dv
←a = vdv/dx*
* These three are tricky.
Substitute original known
back into equation after
differentiation to remove
unwanted variable.
a(v)
→òdt=ò(1/a)dv
←a = dv/dt*
→òadt=òdv
←a = dv/dt
a(t)
→òdt=ò(1/v)dx
←v = dx/dt *
v(x)
x(t)
→òvdt=òdx
←v = dx/dt
v(t)
Examples
1. Sprinting
Given: A sprinter accelerates from rest at a constant 9 𝑚/𝑠 2 for 1.5 seconds before
reaching maximum speed.
Find: What is the maximum speed?
Solution:
1.5
𝑣 = ∫ 𝑎𝑑𝑡 =
9𝑑𝑡 = 9𝑡
0
1.5
0
= 𝟏𝟑. 𝟓 𝒎/𝒔
2. Braking Distance
Given: Given an initial velocity of 60kph and a constant acceleration of −1𝑔 =
−9.81 𝑚/𝑠 2 .
Find: How far does the car go before stopping?
Solution: Acceleration is a constant, and we want to know v(x), so we can start in the
upper left box, a(x), and move to the right, to v(x) using the relationship
∫ 𝑎𝑑𝑥 = ∫ 𝑣𝑑𝑣
Including the limits of velocity of 60kph (16.7m/s) to 0kph, and the limits on displacement
of 0 at the starting point, and the unknown final distance x, we get:
𝑥
0
𝑎𝑑𝑥 =
0
𝑣𝑑𝑣
60
We know acceleration, so we can integrate both sides.
𝑥
𝑥
𝑎𝑑𝑥 =
0
−9.81𝑑𝑥 = −9.81 𝑥
0
0
1
𝑣𝑑𝑣 = 𝑣 2
2
16.7
0
16.7
=
𝑥
0
= −9.81 𝑥 − 0 = −9.81𝑥
1 2
0 − 16.72 = −278.9
2
Thus
−9.81𝑥 = −278.9 → 𝑥 =
278.9
= 𝟐𝟖. 𝟒 𝒎
9.81
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3. Pendulum
a. Given: The horizontal position of a pendulum is given as 𝑥 𝑡 = 3 cos t [m]
b. Find: The maximum acceleration.
c. Solution:
We see that we must differentiate twice to get from x(t) to a(t):
𝑑2
𝑎 𝑡 = 2 𝑥 𝑡 = −3 cos 𝑡 [𝑚/𝑠 2 ]
𝑑𝑥
But at what time is the acceleration maximum? This is an important point, so think this
over carefully. In this case, the acceleration is at maximum when velocity is zero. We
could plot sine and cosine curves to show you graphically, but it might be more helpful to
simply imagine yourself on a swing, being pushed very high up. Your stomach will churn
most not at the bottom, where velocity is high, but at the extremes of height, where you
stop to swing back in the other direction. To find when velocity is zero, we can
differentiate x(t) once:
𝑑
𝑣 𝑡 =
𝑥 𝑡 = −3 sin 𝑡 [𝑚/𝑠]
𝑑𝑥
And then set velocity equal to zero, and solve for the time:
𝑣 𝑡 = 0 = −3 sin 𝑡 → 𝑡 = asin 0 → 𝑡 = 0, [𝑠]
Of course there will be many times when v = 0, but we only need one. Substitute this
time into the acceleration equation to find out what the maximum acceleration is:
𝑎𝑀𝑎𝑥 = 𝑎 𝑡 = 0 = −3 cos 0 = −3 𝑚/𝑠 2
Of course, with another time where v=0, we could find that the value of 𝑎𝑚𝑎𝑥 is positive.
It depends on the direction the pendulum is swinging.
Relating Force and Motion
There are several equally fundamental ways to relate force and motion. They are:
1. Newton’s Second Law (Force equals mass times acceleration)
2. Tracking the energy added, removed, and stored in the system as it moves a certain distance.
3. Tracking the change in momentum of a system over time
Pay careful attention to the wording above. Momentum and time are related, as are energy and distance.
Newton’s second law is the only one that incorporates acceleration directly. We saw in the previous
section that all these parameters can be related, so it shouldn’t be surprising that we say that these
approaches are equally valid. Based on the known quantities, and the goals of the analysis, one of these
methods may be more direct than another. All the methods require knowledge of force and mass, but
the parameters of motion are different. This is summarized in the table below:
Method
Acceleration
F=ma
X
Energy
Momentum
Velocity
Displacement
X
X
X
Time
X
Proficient use of these methods will require more time than we have, and much more practice, but we
will solve a single problem with each method, just to demonstrate the idea.
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Example – Box Being Pushed
Given: a 3kg box at rest on a flat conveyor belt. The belt starts to move, and a friction force of 5N is
applied to the box.
Goal: Apply each of the methods to this problem, and see what information they provide.
The first step with any of these methods (and most engineering problems!) is to draw a “Free Body
Diagram” (FBD), which we first encountered when discussing Statics The picture above does a good job
showing the situation, but the forces and masses aren’t obvious. A FBD is an idealized sketch of the
problem, and has all forces and masses clearly labeled. It is a useful stepping stone between a physical
problem and a mathematical equation.
𝐹𝑤𝑒𝑖𝑔 ℎ𝑡
y
x
𝐹𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 10𝑁
Method 1: Newton’s Second Law
Review of the Method
The mathematical expression of Newton’s Second Law should be familiar from physics class. The
acceleration of a body is proportional to the force, and inversely proportional to the mass.
𝐹 = 𝑚𝑎
Example
We can apply Newton’s 2nd Law in order to find the acceleration of the box in the x direction. Note that
we include a summation sign. If there were more than one force in the x direction, they would simply be
added, just as we say in statics:
Σ𝐹𝑥 = 𝑚𝑎𝑥
10 𝑁 = 3 𝑘𝑔 ⋅ 𝑎
10 𝑁
10 𝑚
𝑎=
=
3 𝑘𝑔
3 𝑠2
Using the techniques of the previous section, we could now find velocity as a function of time or distance,
but extra steps run counter to the engineering ethos of optimization of effort. If we need to know
velocity, distance or time, it might be easier to use another method.
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Method 2: Energy and Work
Review of the Method
The mathematical expression Kinetic Energy (K) should be familiar from physics class.
1
𝐾 = 𝑚𝑣 2
2
The quantity of “Work” as used by engineers is equal to the magnitude of an applied force, times the
distance the part moves:
Work = Force × Distance
Work and energy are related by the following equation:
𝐾1 + 𝑊1→2 = 𝐾2
Where 𝐾1 and 𝐾2 are the kinetic energy at the beginning and end of the move, and 𝑊1→2 is the work
done as the part moves over a certain distance between points 1 and 2.
Example
Continuing with our example of the box on the conveyor belt, we weren’t really given enough info about
the problem. Let’s say that we want to find out the velocity after the box has moved 5cm. We can start
the solution by finding the kinetic energy at x=0 and x=5cm:
1
1
𝐾1 = 𝑚𝑣12 = 𝑚 0 2 = 0
2
2
1
1
𝐾2 = 𝑚𝑣22 = 10𝑘𝑔 𝑣22 = 5𝑣22 𝐽
2
2
And the work done on the box will be:
𝑊1→2 = 𝐹 ⋅ 𝑑 = 10𝑁 ⋅ 0.05𝑚 = 0.5[ 𝐽]
Substituting these values into the governing equation gives:
0 + 0.5 = 5𝑣22
𝑣2 =
0.5
𝑚
= 0.31
5
𝑠
There are several important things we should emphasize. First, we never needed to find acceleration.
Second, we never needed to consider any vectors. This may not be as apparent with such a simple
example. Work and energy are both scalars, and so we did not need to be concerned with vector algebra.
That is to say, the math is relatively simple, even for more complicated problems.
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Method 3: Impulse and Momentum
Review of the Method
The quantities of impulse and momentum are likely familiar to you, but it is likely that you have not seen a
rigorous mathematical definition of them before. The definition of momentum aligns closely with the
everyday use:
Momentum = Mass × Velocity
The higher the mass of an object, and the higher the speed, the less you want to get in its way. For
example, Indiana Jones is rightly afraid of the giant rolling stone. It has both high mass and high velocity.
Impulse is simply the change in momentum over time, which is the same as integrating the applied force
over time. It can be calculated as
𝐼𝑚𝑝1→2 =
𝑡2
𝐹𝑑𝑡
𝑡1
The momentum at time 1 and time 2 are related by an equation that is very similar in form to the
equation for work and energy:
𝑚𝑣1 + 𝐼𝑚𝑝1→2 = 𝑚𝑣2
Where 𝑚𝑣1 and m𝑣2 are the momentum at the beginning and end of a time interval, and 𝐼𝑚𝑝1→2 is the
impulse imparted on the part between time 1 and 2.
Example
Returning for a final go on our example of the box on the conveyor belt, again, we were not given enough
info about the problem. Let’s say that we want to find out the velocity after the belt has been moving for
0.5 seconds. We can start the solution by finding the momentum t=0 and t= 0.5 sec:
𝑚𝑣1 = 3[𝑘𝑔] 0 2 = 0
𝑚𝑣2 = 3 𝑘𝑔 𝑣2 = 3𝑣2
And the impulse applied to the box by friction with the conveyor belt will be:
𝑡2
0.5
𝑘𝑔 𝑚
𝑊1→2 =
𝐹𝑑𝑡 =
10𝑑𝑡 = 10 𝑡 0.5
=5
0
𝑠
𝑡1
0
Substituting these values into the governing equation gives:
0 + 5 = 3𝑣2
5𝑚
𝑣2 =
3𝑠
There are several important things we should note. First, we never needed to find acceleration, or know
anything about the distance travelled. Second, it is very important to note that momentum and impulse
are vectors, and while this problem is very simple, care must be taken to keep track of direction in more
complex problems.
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It might be interesting to see how we can find the momentum-impulse balance starting with Newton’s
Second Law.
𝐹 = 𝑚𝑎
Using the definition of acceleration, we can put this in terms of velocity and time.
𝑑𝑣
𝐹=𝑚
𝑑𝑡
If we assume that mass is constant, we can bring it inside the derivative.
𝑑 𝑚𝑣
𝐹=
𝑑𝑡
We can move the “dt” to the other side of the equations
𝐹 𝑑𝑡 = 𝑑 𝑚𝑣
integrate this to get:
𝑡2
𝑡1
𝐹 𝑑𝑡 = 𝑚𝑣
𝑣2
𝑣1
= 𝑚𝑣2 − 𝑚𝑣1
Finally, rearrange the above equation to get the balance we expected.
𝑚𝑣1 +
𝑡2
𝑡1
𝐹 𝑑𝑡 = 𝑚𝑣2
Conclusion
Relating acceleration velocity time and position allows you to manipulate any knowledge you may have
about a problem to either answer a question directly, or put it into a form that is easier to use. It is not a
theoretical exercise. The three ways we discussed to relate force and motion, Newton’s 2nd Law, Work
and Energy, and Impulse and Momentum, can all be applied to any problem. An important question for
the practicing engineer is to determine which is most appropriate for the job at hand.
8|Dynamics