Dynamics Lectures 1 & 2

7/29/2019 Dynamics Lectures 1 & 2

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Dynamics

This document represents part of the lecture notes for the course ENGR243-Dynamics given at

Concordia University, on winter 2012. It should be used as a complement to the Textbook,

which is Vector Mechanics for Engineer: Dynamics, ninth SI edition, by Beer and Johnston (or

some equivalent editions, 3,5, 6, or 7). Some other documents have been used to prepare those

lecture notes:

-Dynamics, by Lawrence E. Goodman & William H. Warner, Dover, 2001

-Engineering Mechanics, By McLean and Nelson, Schaums outline series, 1962

-Classical Dynamics, by Marion, Thornton,


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Chapter 1: Kinematics

Let us start the study of Dynamics by the introduction to the subject of Kinematics. Kinematics

is the part of physics that study the motion of body, either particle object or rigid body, butwithout seeking to find the cause of motion. This means, we will be looking at the motion only,

but not at what are the conditions that must be reach so that the motion happen.

Part 1: Kinematic of a particle

Definitions and basic concept

Let us present the first initial definition needed to proceed to the correct study of kinematics.

The initial definition needed is the position of a particle. If one is working in a Cartesian one-

dimensional world, then:

the position is the relative distance from a point called the origin, relative to the unit

used.

In SI units, the unit of distance is the meter. It has to be specified (or at least, you should

indicate that you are working in SI units). Figure 1 show a point particle at position x = 3 m on

a given axis.

Figure 1: position on a 1 dimensional axis.

Now, there are two definitions that one must understand in order to proceed. The first one is

the displacementand the other one is the traveleddistance. The displacement is the most

useful definition in application, although the traveled distance is of common use.


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The displacement is the variation of position through time.

The traveled distance is the total distance traveled between two times coordinate.

One calculate the displacement as

(1.1)There is no simple explicit formula for the traveled distance, since it is a function of the actual

path taken by the particle. The main difference between the two definitions can be outline with

the following example:

Example 1: Find the displacement and the traveled distance for the path in figure 2.

Figure 2

The displacement: x = 5 3 = 2m

The traveled distance: from 3m to 1 m: +2 m

From 1 m to 7 m: + 6 m

From 7 m to 5 m: + 2 m total: 10 m

Now, the word time was introduced for the definition of displacement. The reason is that in

physic, there is no instantaneous motion. SO, to go from the initial position to the final position,

a certain laps of time has occurred. The displacement divided by the timed elapsed to go from I

to f is called the mean velocity.

(1.2)

The instantaneous velocity, also called the velocity, is the mean velocity taken between twoinfinitely small time periods.

(1.3)


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A small note is important to add here: the velocity is a vector. The notation has been dropped

for simplicity since the definitions are given in 1 dimension. The correct notation will appear in

the next section.

Of course, the velocity is not a constant for eternity! It can change with time. We define the

mean acceleration as in 1.4 and the instantaneous acceleration (acceleration) as in 1.5. Again,

those values are vectors. A small discussion on this fact in 1D will follow:

(1.4)

= = (1.5)

Also, one can have this alternative definition of the acceleration:

=

(1.6)

Exercise: try to prove that

Discussion on the vector meaning in 1 D:

The position can be either positive or negative, the velocity and the acceleration as well. Now,

the meaning of a negative position is a particle in the negative axis according to the arbitrary

system of axis. A velocity in the negative is indicating that the particle is moving towards the

negative. It indicates the direction of the velocity vector. Similarly, a positive acceleration

indicates that the velocity is getting larger and larger in the positive and a negative acceleration

means that the velocity is getting larger and larger in the negative values.

What was presented so far are the definition of the basic concepts. One can ask himself (or

herself) if there is a relation other than the derivative between those quantities. The first

thing to notice is the fact that one can express the velocity as a function of the position: v(x).

Also, the acceleration can be written as: a(x). But it can also be written has: a(v). Now, since we

have function, it is possible to invert then, ie, write x(v), x(a), v(a), and even utilize the value of

time and write: x(t), v(t), a(t).

In other words, one needs to be able to play around equations to completely describe system

using those definitions.

Important remarks:

1- If one write x(t), one can use: 2- The same can be found for the velocity:


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3- Similar integral equation may be found when working with those function (see theexamples)

4- The following table give some variables that will be used in kinematics and where one isexpected to find them:

Linear rotation

time t, in second (s) t, in seconds (s)

position x, y, z, r, s, in meter (m) , in radian

velocity v, , , , , , acceleration , , , , , , , ,

Example 1: The position of a particle is given by 3 + 4 + 2 3 , where x is inmeter. Find:

a) the time at which the velocity is zero

b) the acceleration at t = 4s.

Solution:

a) The first thing to do is to find the value of the velocity. According to 1.3: 3 + 4 + 2

3 4 + 4 9

So, if v = 0, then one has a quadratic equation to solve:

0 4 + 4 9 imply the use of : with a = -9, b = 4, c = 4So the two possible values of times are: t = 0.9249 s and t = -0.4805 s. The second time is

usually rejected as the majority of the physical problem studied in dynamics starts at t =0s.

b) Let us find the acceleration (1.5) 4 + 4 9

4 18 At t = 4s, one finds an acceleration of -68 m/s

2.


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Example 2: Let a particle have the following acceleration, k being a constant: . If the initialconditions specify that at x = 0, v = 1 m/s and at x = 1 m, v = 3 m/s, find the value of the constant. Also,

find the velocity if x = 2 m

Solution:

Let us find an expression for the velocity as a function of the position (according to the initial value

given, it is the way to go). So, from 1.6:

7+

Let us isolate the v on one side and then integrate:

7 + 7+

ln3 = 87 = ln3ln 87= 8.2274

For the velocity if x = 2m:

=

.

So:

ln ln1 = ln = 2.0677Therefore, the velocity is v = 7.9066 m/s.

All the above discussion is quite interesting, but an important fact was omitted: the importance of the

reference system. So far, only one reference system was used. What if there is more than one system of


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reference? For example, a person standing still on the side of the road, versus a person in a bus driving

by. Those are two different reference system, if, one consider the corner of the street as the origin (the

standing still reference) and one consider the end of the bus as the origin (the moving reference).

If one consider the fixed system as x and the moving one as x, than a position can be different

depending on the observer. This fact can be related to what is called the Galilean relativity (from adiscussion by Galileo Galilei, in is famous discussion on two world system). In figure 3, the position of a

particle is given according to each system. If one consider the difference between the two system as

/, then:

= / (1.7)

Figure 3: A picture of the two references systems.

Of course, one can go one step further and find the value of the velocity and of the acceleration:

= / (1.8) = / (1.9)

Those equations are called the relative velocity and relative acceleration. They are very useful whendealing with systems of many objects. For example, with one system of axis, if there are two particles, a

particle A at xA and a particle B at xB, relation 1.7 to 1.9 still hold.

Figure 4: a) Atwood machine and b) (the left part), mass and pulley system.


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A lot of physical systems use more than one object. For example, one may think of a simple Atwood

machine (figure 4a)), made of two masses and a pulley. The displacement of a mass of the system leads

automatically to a displacement of the other mass. For the case of figure 4b), one can write that the

length of the cable holding together the mass is a constant. So, one has equation 1.10. The first and

second derivative over time of that equation leads to equation 1.11 and 1.12, which are examples of

constraint movement.

2 = (1.10) 2 = 0 (1.11) 2 = 0 (1.12)

Let us now look at some particular cases of one dimensional kinematic:

1- Uniform Rectilinear MotionThe uniform rectilinear motion is a special case where there is no acceleration. Hence, a = 0. So,

using the definition of the acceleration, one gets a constant velocity (first integral of the

acceleration) and a linear function for the position as a function of time.

= (1.13)= (1.14)

Even if such a motion is very simple, it has practical application that will be put forward in the 2

dimensional system.

2- Uniformly Accelerated Rectilinear MotionLike the name says, the acceleration is a constant. Let us try to find the value of the velocity (in

function of time and in function of the position) and of the position.

Velocity:

Let us start with the standard definition of the acceleration:

=

In separating the dt of the dv and integrating on each side, one has:

= (1.15)But, if one start with the other definition of the velocity (equation 1.6), then:

= = 2

2 =


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= 2 (1.16)Position:

If one goes back to equation 1.15, and replaces the value of v by the derivative of the velocity over time,

one gets:

= =

So, integrating the last term and re-writing the result in a proper fashion (left in exercise), one gets:

= (1.17)Equations 1.15 to 1.17 are the equations of the uniformly accelerated motion. Let us look at the most

famous application:

Free Fall of a body

In the free fall of an object near the surface of the Earth, the acceleration is near a constant and is

supposed to be -9.81 m/s2. This means that all of the objects falling towards the ground, near the

ground, obey the equation 1.15 to 1.17. Of course, it is an approximation since the resistance coming

from the air is neglected. But, the approximation is still good.

This result was reported by Galilee in the famous tower of Pisa experiment. Although historians are not

sure if it really happens, the legend says that Galileo dropped two balls from the top of the leaning

tower (and yes, it was leaning at that time). The two balls were of different weight. Legend says that,

according to the prediction, the ball touched the ground at the same time. The result was going against

what was believed to be true at that time, so it was revolutionary.

Example 3:

A particle is moving in the x direction with a constant acceleration of 4 m/s2. If the initial velocity was of

-3 m/s, what is the displacement of the particle between time 1 s and time 4 s?


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Solution:

Using 1.17, with t0 = 1 s, the problem can be very fast. But, to do so, one need to find v0, which in that

case becomes the velocity at time 1s. So, let us use 1.15 first:

+ = 3 + 4 1 0 = 1 Now, with 1.17:

= = + 2 = 1 4 1 + 2 4 1So, x=21 m.

Example 4:

An extreme diver falls from a cliff 26 m high in the sea below. At what speed does he touch the water?

Solution:

This is a problem of free fall. Let us indicate all the variables that we have:

9.81 26 0 0 0

It is often useful to indicate the variables in a schematic of the problem.

In this problem, the velocity is needed, so the use of equation 1.16 is a good choice, since the value of

the position and the initial velocity are known.

0 + 2 9.81 0 26 22.59 81.31 !

Kinematics in more than one dimension

Let us now focus on kinematic of a particle in more than one dimension, mainly in two dimensions, but

also to some extent, in three dimensions. The key to be able to jump to higher dimension is to be

capable of playing with vectors. Therefore, let us introduce the vector version of the position, the

velocity and the acceleration.

Let one consider the Cartesian space of figure 5 as the reference system to be used. In that figure, one

can find a particle at a particular position called P, with the coordinate (x,y,z). The position can be

represented by a position vector: . In Cartesian coordinate, this vector is generally indicated by itscomponents in x, y and z.

+ + (1.18)


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Figure 5: 3-dimensional Cartesian space.

Now, it is possible to go back to the definition presented at the beginning of the kinematic section and

find some expressions for the velocity and the acceleration. Using vector calculus, one finds:

+ +

+ + (1.19)And for the acceleration:

+

+

(1.20)A more complicated case appears when one seek the acceleration of a particle, but when the velocity is

a function of the position as oppose to a function of time: , , . Sometimes, the velocity is evenmore complex, as being a function of both the position and time for every component (as in fluid

dynamics). In this case, the acceleration becomes:

+

+

+

(1.21)

In 1.21, the dot notation has been used to indicate the time derivative and the chain rule for derivation

have been applied. Of course, similar expression can be found for the y component and the z

component. It is to note that equation 1.21 is a generalization of the equation 1.6.

All the above discussion is based on the Cartesian coordinate system. This system is very useful, but it is

not the only one available. Let us present the normal/tangent system and the radial/transverse system.

Those two systems are in 2 dimensions and they are very useful when dealing with curvilinear motion

and rotation. In 3 dimensions, the most common systems used are the rectangular (Cartesian), the

cylindrical (3D polar) and the spherical system.

Normal/Tangent:

Let us consider a system of two perpendicular axes fixed on a moving particle. We define one of the axes

to be along the tangent to the path of the particle (see figure 1.6). It is possible to put a unit vector onthe tangent axis. Of course, this vector is parallel to the velocity of the particle. So, the velocity, using

this type of system, is always on the tangent axis.


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Now, if there is a change in the direction of the velocity, the axis will change direction as well (see also

figure 1.6). Let us form a triangle with the two unit vector, separated by an angle . The third vector

closing the triangle represents the change in the direction of the tangent unit vector. By geometry:

2 sin 2 , lim

= 1

This implies that the derivative of the unit tangent unit vector is a unit vector, whose direction is toward

the change (the center of rotation). This vector is called the normal unit vector, (see figure 1.7).

Figure 1.6: tangent vector and variation of the tangent vector

Figure 1.7: The representation of the normal/tangent system

The position in such system is simply the position, s, on the path of the particle. It is assumed that the

path is the only place where the particle may reside. The velocity, like mentioned before, is entirely on

the tangent axis.

= (1.22)The acceleration is a bit more complicated:

= = + Now, in the last term, the chain rule indicates that there is a multiplication of three other values. The

first derivative is the definition of the normal unit vector. The second one is the inverse of the radius of

curvature at a particular point ds (use the definition of an angle in radian) and the last derivative is the

velocity along the path, v. So:

= + (1.23)


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The at is the tangential acceleration, mainly the change in the magnitude of the velocity in time. The

other term, often called an or ac is the normal acceleration (centripetal acceleration), and it represent

the change in direction of the velocity in time. It is always directed toward the center of curvature.

*it is possible to extend this system in 3D, with the binormal vector eb , form by the cross product of the

tangential and normal vector. In that case, the plane made by e t and en is called the osculating plane. Itis seldom useful, except in some particular torsional motion.

Example:

a) If a particle has a velocity of 5 m/s and an acceleration of 5 + 5 m/s2, find the radius ofcurvature.

b) Repeat the question if 2+ 5and 3+ 4.Solution:

a) We know that the radius appears in the normal acceleration. So: 5

It is given that v = 5 so: 5 b) Let us find v: 4 + 25 5.385

Now, what to do with the acceleration? Let us find the tangent acceleration by projecting the

acceleration in the direction of the velocity:

25.385 3 +5

5.385 4 4.828So, the normal acceleration is:

5 + 4.828 1.3 Hence, te value of the radius of curvature is 22.306 m

Radial/Tranverse

The system radial/transverse is sometimes called the polar system. Figure 1.8 represents the system in a

2D motion, related to the xy axes. The two unit vectors are: . In the present case, a usefulrelation exists between the two vectors:


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Figure 1.8: schematic of the system used

Now, let us write the value for the position, the velocity and the acceleration. The calculation is left to

the reader. Hint: use the chain rule and the expression given above.

(1.24) + (1.25)

+ 2 + (1.26)

Those equations will be interesting in many cases of rotating object. In the simplest case, one canimagine a rotation on a circle, so r is a constant.

Example:

A particle travel on a spiral path 2 ., with = 1.5 , a constant. Find the velocity and theacceleration when = 7 /6.Solution:

Cylindrical system:

The cylindrical system is a system based on the polar coordinates in 2D, with a vertical distance from the

plane made by the polar coordinates. So, a position is normally given by:

, ,


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To use such coordinate, it is of usage to either identify the value of r, and z to the x, y and z rectangular

components, or to introduce new units vectors. Let us consider the triad of vectors , , , aspresented in the figure 1.9. Hence, it is possible to write (the demonstration of that is left to the reader):

= =

= 2 +

Figure 1.9: representation of the cylindrical coordinates.

Spherical system:

A useful system of axes for the spherical symmetric system is the spherical system. The unit vectors for

such a system are presented in figure 1.10. It is a system made by a triad moving on a tennis ball. For

this system, one uses the triad of vectors , , . The relations with the rectangular system ofaxes are:

= cos , = sin , = The position, the velocity and the accelerations are: (see Goodman and Warner for details)

= =

= 2 + + 2 + 2

Figure 1.10: Spherical system


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Relative motion:

A quick discussion on relative motion can be presented here. Has was discussed above with the Galilean

relativity, the same equations apply in 3 dimensions. Hence, if one takes a system A in relative motion to

a system O, fixed or having a constant velocity, one has:

+ / (1.27) + / (1.28) + / (1.29)

A visualization of the motion is presented in figure 1.11.

Figure 1.11: The relative motion

Application: 2 D projectile motion

The particular case of the projectile is a very useful one in engineering. It is a 2D motion representing an

object with (or without) an initial velocity, under the influence of the gravitational acceleration only. So,

the accelerations in x and in z are taken to be zero (there is no motion in z). The acceleration in y is the

constant for the gravitational acceleration near the Earth surface: -9.81 m/s2. The minus sign is only to

indicate that it is directed towards the ground.


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In such a motion, since the acceleration in x is zero, it is a rectilinear uniform motion in that direction.

Since the acceleration in y is non-zero but constant, it is a uniformly accelerated motion. The

combination of both motions gives a curvilinear motion in xy (a parabola). The equations governing such

motion can readily be written from the 1D description:

+ (1.30) + 4.905 (1.31)

9.81 (1.32) 19.62 (1.33)

How to use those equations:

1- Identify all the given information, and draw a schematic2- Clearly state what is asked for (the missing information)3- Find, in the four equations, the one or the ones that can help you4- Solve

Those steps must be applied to every problem. It is a good way to keep a good view of what you are

doing.

Example:

A ball is thrown from a high of 2 m, at a velocity of 15 m/s at an angle of 30 from the horizontal plane.

At what distance from the origin does the ball land?

Solution:

One has the following initial values: y0 = 2m , y = 0 , x0 = 0 m, vx0 = 15cos(30) , vy0 = 15sin(30) , t0 = 0

One has to find the value of x. From equation 1.30, we have:

So, if one can find the time t, one has x. By 1.31:

0 2+ 7.5 4.905


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So, t has 2 possible MATHEMATICAL value of: t1 = -0.232s and t2 = 1.761s.

The physical value, according to this problem is t = 1.761 s, so x = 22.876 m.

*Kinematic of rotation:

It was mentioned before that the kinematic of rotation is the same thing as the kinematic without

rotation, in a general manner. In a more specific manner, if one look at the Uniformly Accelerated

Circular Motion (UACM), the following equations are useful. They will be very useful in the Rotation of

the rigid body. Let us denote the radius of rotation by r, the distance in meter by s (circular motion), the

angular position by , the angular velocity by and the angular acceleration by :

= ; = ; =

= + + 2

= + = + 2

Dynamics Lectures 1 & 2

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