05 Curvilinear Motion Concept Overview
PARTICLE CURVILINEAR MOTION | CONCEPT OVERVIEW
The TOPIC of PARTICLE CURVILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: Whereas in RECTILINEAR MOTION a PARTICLE undergoes MOTION in a STRAIGHT LINE, CURVILINEAR MOTION describes a PARTICLE that is moving along a CURVED PATH. This CURVED PATH of MOTION can fall within a TWO or THREE DIMENSIONAL SPACE. At the foundation of analyzing a PARTICLE undergoing CURVILINEAR MOTION is a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE that gives definition regarding the POSITION of a PARTICLE at a particular point in time using POSITION VECTORS.
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RECTANGULAR COORDINATES: When working within a RECTANGULAR SPACE, we are operating within a standard CARTESIAN COORDINATE SYSTEMβ¦one that we are all very familiar with, illustrated as:
A CARTESIAN COORDINATE SYSTEM gives us a means to specify each point uniquely within a plane using a PAIR of COORDINATES for TWO DIMENSIONAL spaces, or THREE COORDINATES in a THREE DIMENSIONAL space. These individual COORDINATES are assigned a distance from a DIRECTED LINE, referred to more commonly as a COORDINATE AXIS, measured in some established unit of measurement. The point where all AXES meet simultaneously is referred to as the ORIGIN, located at (0,0) in a TWO DIMENSIONAL space and (0,0,0) in a THREE DIMENSIONAL space.
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POSITION: A PARTICLE with the COORDINATES of (π₯, π¦, π§) will be represented by a POSITION VECTOR, π, drawn from the ORIGIN to the location of the PARTICLE.
This POSITION VECTOR represents the THREE COMPONENTS, or DIMENSIONS, of the PARTICLE, expressed in VECTOR NOTATION as: π = π₯π + π¦π + π§π The CURVED PATH represents a SEQUENCE of POSITIONS for the PARTICLE and is a FUNCTION OF TIME. Each COMPONENT is also operating as a FUNCTION OF TIME, which allows us to define where in space precisely it resides as it relates to a particular axis in our rectangular frame of reference.
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If over a period of TIME, this PARTICLE changes POSITION within the same established REFERENCE FRAME, having new COORDINATES (π₯β², π¦β², π§β²), its POSITION can be represented by a new POSITION VECTOR, πβ², such that: πβ² = π₯β²π + π¦β²π + π§β²π The change in POSITION of this PARTICLE is expressed as: βπ = π 0 β π Or: βπ = (π₯ 0 β π₯)π + (π¦ 0 β π¦)π + (π§ 0 β π§)π This change in POSITION is more commonly referred to as the DISPLACEMENT, which as you can see, is also represented as a VECTOR.
VELOCITY: Given a period of time starting at π‘3 and ending at π‘4 , the AVERAGE VELOCITY of the particle can be modeled as the CHANGE IN POSITION OVER THE CHANGE IN TIME, or in formulaic terms:
π£678 =
π4 β π3 π‘4 β π‘3
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Otherwise written as:
π£678 =
βπ βπ‘
Where: β’ βπ is the DISPLACMENT β’ βπ‘ is the PERIOD of TIME this DISPLACEMENT takes place As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, expressed as:
π£=
ππ ππ‘
In this case, the POSITION VECTOR, π3 and π4 , of the PARTICLE are so close that they can be considered to fall on the same line, and the very small ππ that is represented in the formula for INSTANTENOUS VELOCITY is a VECTOR that runs PERPINDICULAR to these TWO POSITION VECTORS. In other words, the INSTANTENOUS VELOCITY is a VECTOR that runs TANGENT to the CURVED PATH of MOTION at this particular point. So the lesson to remember here is that the DIRECTION of the INSTANTENOUS VELOCITY is always TANGENT to the CURVED PATH.
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VELOCITY is the first TIME DERIVATIVE of POSITION and can be represented in VECTOR TERMS as: π£ = π₯π + π¦π + π§π Remember that when we are cruising through the NCEES Reference Handbook, that the dots that fall above the x, y, and z variables are specifically defining those variables as TIME DERIVATIVES, or otherwise: ππ₯ = π£: ππ‘ ππ¦ π¦= = π£; ππ‘ ππ§ π§= = π£< ππ‘ π₯=
This shows us that we can also write the VELOCITY vector in terms of its VELOCITY COMPONENTS, such that: π£ = π£: π + π£; π + π£< π All these expressions are EQUIVALENT.
ACCELERATION: ACCELERATION is defined as the RATE OF CHANGE in the VELOCITY of a PARTICLE in motion.
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It can be more memorably referred to as the TIME derivative of VELOCITY, or in SCALAR FORMULAIC TERMS:
π=
ππ£ ππ‘
It may also be referred as the SECOND TIME DERIVATIVE of the POSITION: π4 π π= 4 ππ‘ Represented in VECTOR TERMS as: π = π₯π + π¦π + π§π In this case, the double dots that fall above the x, y, and z variables are specifically defining those variables as SECOND TIME DERIVATIVES, or otherwise:
π₯=
π4 π₯ = π: ππ‘ 4
π4 π¦ π¦ = 4 = π; ππ‘ π4 π§ π§ = 4 = π< ππ‘ These expressions will indeed come across as intimidating, but with a small amount of review and application, you will find that they are no more difficult to work with than their SCALAR counterparts.
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Using the definitions stated, we can also express the ACCELERATION VECTOR in terms of its COMPONENTS as: π = π: π + π; π + π< π Unlike its VELOCITY counterpart, the DIRECTION of the ACCELERATION VECTOR, in general, does not run TANGENT to the CURVED PATH of the PARTICLE.
POLAR COORDINATES: Up to this point, the method that we have used to define the POSITION of PARTICLE undergoing CURVILINEAR MOTION is that of using a standard CARTESIAN COORDINATE SYSTEM which gives us a means to define the POSITION using simple RECTANGULAR COMPONENTS. This is a SYSTEM and METHOD of measurement that we are most familiar and comfortable with. In a TWO DIMENSIONAL SPACE, this system utilizes a HORIZONTAL X-AXIS and a VERTICAL Y-AXIS. In a THREE DIMENSIONAL SPACE, a THIRD AXIS is added to provide a means to MEASURE the DEPTH, this axis being the Z-AXIS. However, there is a second, equally important set of COMPONENTS and COORDINATE SYSTEM that we must dive in to, that being, POLAR COORDINATES and the CYLINDRICAL COORDINATE SYSTEM.
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Setting up a standard CYLINDRICAL, or POLAR, SPACE allows us to define the position of a particle at any point of its path using position vectors. As we did in the THREE DIMENSIONAL CARTESIAN SPACE, we have a vertical ZAXIS and replace the PLANAR X and Y-AXIS with our RADIAL and TRANSVERSE POSITION measurements, π and π: The NCEES doesnβt go in to using the THIRD DIMENSION in a CYCLINDRICAL SPACE, so we will forgoe discussion on it as well to avoid any confusion and further time in reviewing the material.
POSITION: The POSITION of any particle, is represented by a POSITION VECTOR from the origin to the location of this PARTICLE which can be expressed as: ππ = ππB NOTE, the NCEES does not clearly delineate the difference between r and r as it is presented in the Reference Handbook, therefore, we have chosen to denote one as ππ which represents the POSITION VECTOR of the PARTICLE whereas, r, is the RADIAL MEASUREMENT along the RADIAL AXIS.
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VELOCITY: In a POLAR COORDINATE SYSTEM, the VELOCITY is still the TIME DERIVATIVE of the POSITION VECTOR, such that:
π£=
πππ ππ‘
It has TWO COMPONENTS along the RADIAL and TRANSVERSE directions, presented in the NCEES Reference Handbook as: π£ = πππ + ππππ½ Where: π = π£B = TIME DERIVATIVE of the POSITION VECTOR ππ = π£E = POSITION VECTOR times the ANGULAR VELOCITY (π), which is the TIME derivative of the TRANSVERSE COMPONENT, π. Knowing these definitions allows us to write an equivalent expression as: π£ = π£B ππ + π£E ππ½
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ACCELERATION: As it is in the RECTANGULAR COORDINATE SYSTEM, the ACCELERATION is defined as the RATE OF CHANGE in the VELOCITY of a PARTICLE in motion, or in other terms, the TIME derivative of VELOCITY:
π=
ππ£ ππ‘
Or the SECOND TIME DERIVATIVE of the POSITION VECTOR: π4 π π= 4 ππ‘ It has TWO COMPONENTS along the RADIAL and TRANSVERSE directions, presented in the NCEES Reference Handbook as: π£ = (πβππ 4 )ππ + (ππ + 2ππ)ππ½ Where: (πβππ 4 ) = πB = π and is the second TIME derivative of the POSITION VECTOR ππ + 2ππ = πE = π and is the second TIME derivative of the TRANSVERSE COMPONENT, π, and is also known as the ANGULAR ACCELERATION Knowing these definitions allows us to write an equivalent expression as: π = πB ππ + πE ππ½
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LOCATING A POINT USING POLAR COORDINATES: Given a POINT, P, as illustrated:
To identify the POINT P in space, as illustrated in our rectangular coordinate system, we start from the origin (0,0) and go over to the right 2 UNITS and up 2 UNITS. This gives us the RECTANGULAR COORDINATES of POINT P as (2, 2). POLAR COORDINATES represents another way to locate a point in CYLINDRICAL COORDINATE PLANE. Using the same POINT P illustrated, we could start from the HORIZONTAL AXIS and rotate around the ORIGIN in a COUNTERCLOCKWISE DIRECTION a certain ANGLE of π. By CONVENTION, the HORIZONTAL POSITIVE X-AXIS and a ROTATION in a COUNTERCLOCKWISE DIRECTION constitute the starting point and positive rotation when working within a CYLINDRICAL COORDINATE SYSTEM.
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If we were to ROTATE CLOCKWISE, starting from the HORIZONTAL POSITIVE XAXIS, then we would get a NEGATIVE value for π. As we ROTATE around the ORIGIN, an IMAGINARY LINE some DISTANCE r from the ORIGIN will hit our POINT at P. This tells us that by ROTATING THROUGH an ANGLE ΞΈ and TRAVELING A DISTANCE r will also plot the POINT P that we have defined. The pair of POLAR COORDINATES can then be stated as (π, π), representing the POINT P in a CYLINDRICAL COORDINATE SYSTEM. NOTE that in POLAR COORDINATES, the ORIGIN, as we have referred to it as, is called the POLE and the X-AXIS is called the POLAR AXIS. Illustrating our POINT P in a POLAR COORDINATE SYSTEM, we have:
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CONVERTING FROM RECTANGULAR TO POLAR COORDINATES: To take any POINT P from a RECTANGULAR COORDINATE SYSTEM and QUANTIFY it in a CYLINDRICAL form, we will need to use a little knowledge from TRIGONOMETRY, and specifically, SOH CAH TOA and the PYTHAGOREAN THEOROM. Recall that SOH CAH TOA is a simple mnemonic that we learned back in our early days of MATHEMATICS to remember the various TRIGONMETRIC RELATIONSHIPS developed using RIGHT TRIANGLE geometry. These TRIGONMETRIC RELATIONSHIPS are defined as: π π¦ = β π π π₯ cos π = = β π π π¦ tan π = = π π₯ sin π =
The PYTHAGOREAN THEOREM also states: π₯4 + π¦4 = π4 For a more in depth review on using the various RELATIONSHIPS derived for SPECIAL CASE TRIANGLES, reference the MATHEMATICS section.
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Other relevant information regarding TRIGONOMETRY can also be referenced under the SUBJECT of MATHEMATICS on page 23 and 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Letβs revisit how we started this section with the RECTANGULAR COORDINATE (2, 2) plotted in a simple TWO DIMENSIONAL RECTANGULAR SPACE:
Letβs put some definition around our illustrated POINT, P, in the context of a POLAR COORDINATE SYSTEM. As stated, since we are going HORIZONTALLY 2 UNITS and up VERTICALLY by 2 UNITS, the RECTANGULAR COMPONENTS of POINT P can be defined as: π₯=2 π¦=2 However, we want to define this same POINT in the context of a POLAR COORDINATE SYSTEM.
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We do this by first recalling the convention of measurement in a POLAR COORDINATE SYSTEM, and second, applying our knowledge of TRIGONOMETRY that we outlined above. Recall that in POLAR COORDINATES, the ORIGIN is called the POLE and the POSITIVE X-AXIS is called the POLAR AXIS. By convention, the POLAR AXIS and a ROTATION in a COUNTERCLOCKWISE DIRECTION constitutes a STARTING POINT and the POSITIVE ROTATION. The pair of POLAR COORDINATES will be stated as (π, π), representing the POINT P in a POLAR, or CYLINDRICAL COORDINATE SYSTEM. The DISTANCE from the ORIGIN to the POINT, or otherwise, the r COORDINATE, is often referred to as the RADIAL POSITION COORDINATE. The ANGLE of ROTATION, π, is referred to as the TRANSVERSE COORDINATE. To begin the CONVERSION from RECTANGULAR COORDINATES to POLAR COORDINATES, we will being with the PYHTAGOREAN THEOREM, which again is stated as:: π₯4 + π¦4 = π4
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We have the X and Y RECTANGULAR COORDINATES, which allows us to get the RADIAL POSITION COORDINATE, r, or the LENGTH of the imaginary line from the ORIGIN to the POINT as: (2)4 + (2)4 = π 4 Rearranging and solving for the RADIAL POSITION COORDINATE, r, we get:
π=
2
4
+ 2
4
Or: π=2 2 Thatβs the first piece of our POLAR COORDINATE. Now we need to determine the TRANSVERSE COORDINATE, π, by using our knowledge of the TRIGONOMETRIC IDENTITIES and our trusted mnemonic SOH CAH TOA, which defines for us: π π¦ = β π π π₯ cos π = = β π π π¦ tan π = = π π₯ sin π =
With all of the x, y, and r COMPONENTS defined, we can use any of these relationships to determine the TRANSVERSE COORDINATE, π.
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Letβs use the SINE relationship:
sin π =
2 2 2
Rearranging and solving for π, we get:
π = sinQ3
2 2 2
Or: π = 45Β° With the RADIAL POSITION COORDINATE, r, and the TRANSVERSE COORDINATE, π, defined as: π=2 2 π = 45Β° We can now express the POINT P in POLAR FORM as: (2 2, 45Β°) It is important to NOTE that in the case of POLAR COORDINATES, the POINT P can also be expressed equivalently as having the COORDINATES: (2 2, 405Β°)
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π (2 2, ) 4 π (2 2, 2π + ) 4 And so on. All we are doing here is going from DEGREE to RADIANS and ADDING a FULL ROTATION, either as 360Β° or 2π.
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PARTICLE CURVILINEAR MOTION | CONCEPT EXAMPLE
The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
PROBLEM: At any instant, the horizontal position of the weather balloon can be mapped in space by the function, π₯ π‘ = 8π‘ ft, where t is given in seconds. If the path of flight can be represented by the formula π¦ =
3 3Z
π₯ 4 , the velocity of the balloon after 2 seconds in
flight is most close to: (ft/s)
A. 8 B. 26.8 C. 33.6 D. 64.2
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SOLUTION: The TOPIC of PARTICLE CURVILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are given a scenario where a weather balloon is released in to the atmosphere, rising at a certain rate with a position that can be mapped using a pair of formulas, as defined in the problem statement: π₯ π‘ = 8t ft π¦=
1 4 π₯ 10
We are asked to determine the VELOCITY of this weather balloon 2 seconds in to its flight. The first step in analyzing any PARTICLE undergoing CURVILINEAR MOTION is setting up context around the motion using a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE. In this problem, we will be working with RECTANGULAR COORDINATES, therefore, we will operate within a standard CARTESIAN COORDINATE SYSTEM with the ORIGIN located at POINT A. This will give us a means to define the POSITION of our PARTICLE at a particular point in time using POSITION VECTORS.
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This POSITION VECTOR represents the TWO COMPONENTS, or DIMENSIONS, of the PARTICLE, expressed in VECTOR NOTATION as: π = π₯π + π¦π The CURVED PATH of the weather balloon flight represents a SEQUENCE of POSITIONS for the PARTICLE and is a FUNCTION OF TIME. Each COMPONENT is also operating as a FUNCTION OF TIME, as we see from the problem statement, which allows us to define where in space precisely it resides as it relates to a particular axis in our rectangular frame of reference. The POSITION COMPONENTS can be expressed as: π₯ = 8π‘ ft π¦=
1 (8π‘)4 10
This allows us to define the POSITION of the weather balloon at any point with the POSITION VECTOR:
π = 8t π +
1 (8t)4 π 10
Through our studies in CURVILINEAR MOTION, we know that the VELOCITY is the first TIME DERIVATIVE of POSITION, and can be represented in VECTOR TERMS as: π£ = π₯π + π¦π
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Remember that when we are cruising through the NCEES Reference Handbook, that the dots that fall above the x and y variables are specifically identifying those variables as TIME DERIVATIVES, or otherwise: ππ₯ = π£: ππ‘ ππ¦ π¦= = π£; ππ‘ π₯=
With this, we can define each TIME DERIVATIVE as: π(8t) ππ‘ 1 π 8t π¦ = 10 ππ‘ π₯=
4
Or: π₯=8 π¦=
128π‘ 10
This allows us to define the VELOCITY VECTOR as:
π£ = 8π +
128π‘ π 10
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At a TIME of 2 SECONDS, we get: π£ = 8π + 25.6π This is the VELOCITY, in VECTOR NOTATION, made up of TWO COMPONENTS, i and j, which are PERPENDICULAR to one another. With this, we can express the overall MAGNITUDE of the balloons VELOCITY at this point in TIME as:
π£=
(8)4 + (25.6)4
Or: π£ = 26.8 ft/s The MAGNITUDE of the VELOCITY of the weather balloon 2 seconds after it is released in to the atmosphere is 26.8 ft/s. The correct answer choice is B. ππ. π ππ/π¬
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The TOPIC of PARTICLE CURVILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: Whereas in RECTILINEAR MOTION a PARTICLE undergoes MOTION in a STRAIGHT LINE, CURVILINEAR MOTION describes a PARTICLE that is moving along a CURVED PATH. This CURVED PATH of MOTION can fall within a TWO or THREE DIMENSIONAL SPACE. At the foundation of analyzing a PARTICLE undergoing CURVILINEAR MOTION is a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE that gives definition regarding the POSITION of a PARTICLE at a particular point in time using POSITION VECTORS.
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RECTANGULAR COORDINATES: When working within a RECTANGULAR SPACE, we are operating within a standard CARTESIAN COORDINATE SYSTEMβ¦one that we are all very familiar with, illustrated as:
A CARTESIAN COORDINATE SYSTEM gives us a means to specify each point uniquely within a plane using a PAIR of COORDINATES for TWO DIMENSIONAL spaces, or THREE COORDINATES in a THREE DIMENSIONAL space. These individual COORDINATES are assigned a distance from a DIRECTED LINE, referred to more commonly as a COORDINATE AXIS, measured in some established unit of measurement. The point where all AXES meet simultaneously is referred to as the ORIGIN, located at (0,0) in a TWO DIMENSIONAL space and (0,0,0) in a THREE DIMENSIONAL space.
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POSITION: A PARTICLE with the COORDINATES of (π₯, π¦, π§) will be represented by a POSITION VECTOR, π, drawn from the ORIGIN to the location of the PARTICLE.
This POSITION VECTOR represents the THREE COMPONENTS, or DIMENSIONS, of the PARTICLE, expressed in VECTOR NOTATION as: π = π₯π + π¦π + π§π The CURVED PATH represents a SEQUENCE of POSITIONS for the PARTICLE and is a FUNCTION OF TIME. Each COMPONENT is also operating as a FUNCTION OF TIME, which allows us to define where in space precisely it resides as it relates to a particular axis in our rectangular frame of reference.
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If over a period of TIME, this PARTICLE changes POSITION within the same established REFERENCE FRAME, having new COORDINATES (π₯β², π¦β², π§β²), its POSITION can be represented by a new POSITION VECTOR, πβ², such that: πβ² = π₯β²π + π¦β²π + π§β²π The change in POSITION of this PARTICLE is expressed as: βπ = π 0 β π Or: βπ = (π₯ 0 β π₯)π + (π¦ 0 β π¦)π + (π§ 0 β π§)π This change in POSITION is more commonly referred to as the DISPLACEMENT, which as you can see, is also represented as a VECTOR.
VELOCITY: Given a period of time starting at π‘3 and ending at π‘4 , the AVERAGE VELOCITY of the particle can be modeled as the CHANGE IN POSITION OVER THE CHANGE IN TIME, or in formulaic terms:
π£678 =
π4 β π3 π‘4 β π‘3
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Otherwise written as:
π£678 =
βπ βπ‘
Where: β’ βπ is the DISPLACMENT β’ βπ‘ is the PERIOD of TIME this DISPLACEMENT takes place As the CHANGE IN TIME becomes SMALLER and smaller, approaching 0, the average velocity becomes the INSTANTANEOUS VELOCITY, expressed as:
π£=
ππ ππ‘
In this case, the POSITION VECTOR, π3 and π4 , of the PARTICLE are so close that they can be considered to fall on the same line, and the very small ππ that is represented in the formula for INSTANTENOUS VELOCITY is a VECTOR that runs PERPINDICULAR to these TWO POSITION VECTORS. In other words, the INSTANTENOUS VELOCITY is a VECTOR that runs TANGENT to the CURVED PATH of MOTION at this particular point. So the lesson to remember here is that the DIRECTION of the INSTANTENOUS VELOCITY is always TANGENT to the CURVED PATH.
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VELOCITY is the first TIME DERIVATIVE of POSITION and can be represented in VECTOR TERMS as: π£ = π₯π + π¦π + π§π Remember that when we are cruising through the NCEES Reference Handbook, that the dots that fall above the x, y, and z variables are specifically defining those variables as TIME DERIVATIVES, or otherwise: ππ₯ = π£: ππ‘ ππ¦ π¦= = π£; ππ‘ ππ§ π§= = π£< ππ‘ π₯=
This shows us that we can also write the VELOCITY vector in terms of its VELOCITY COMPONENTS, such that: π£ = π£: π + π£; π + π£< π All these expressions are EQUIVALENT.
ACCELERATION: ACCELERATION is defined as the RATE OF CHANGE in the VELOCITY of a PARTICLE in motion.
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It can be more memorably referred to as the TIME derivative of VELOCITY, or in SCALAR FORMULAIC TERMS:
π=
ππ£ ππ‘
It may also be referred as the SECOND TIME DERIVATIVE of the POSITION: π4 π π= 4 ππ‘ Represented in VECTOR TERMS as: π = π₯π + π¦π + π§π In this case, the double dots that fall above the x, y, and z variables are specifically defining those variables as SECOND TIME DERIVATIVES, or otherwise:
π₯=
π4 π₯ = π: ππ‘ 4
π4 π¦ π¦ = 4 = π; ππ‘ π4 π§ π§ = 4 = π< ππ‘ These expressions will indeed come across as intimidating, but with a small amount of review and application, you will find that they are no more difficult to work with than their SCALAR counterparts.
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Using the definitions stated, we can also express the ACCELERATION VECTOR in terms of its COMPONENTS as: π = π: π + π; π + π< π Unlike its VELOCITY counterpart, the DIRECTION of the ACCELERATION VECTOR, in general, does not run TANGENT to the CURVED PATH of the PARTICLE.
POLAR COORDINATES: Up to this point, the method that we have used to define the POSITION of PARTICLE undergoing CURVILINEAR MOTION is that of using a standard CARTESIAN COORDINATE SYSTEM which gives us a means to define the POSITION using simple RECTANGULAR COMPONENTS. This is a SYSTEM and METHOD of measurement that we are most familiar and comfortable with. In a TWO DIMENSIONAL SPACE, this system utilizes a HORIZONTAL X-AXIS and a VERTICAL Y-AXIS. In a THREE DIMENSIONAL SPACE, a THIRD AXIS is added to provide a means to MEASURE the DEPTH, this axis being the Z-AXIS. However, there is a second, equally important set of COMPONENTS and COORDINATE SYSTEM that we must dive in to, that being, POLAR COORDINATES and the CYLINDRICAL COORDINATE SYSTEM.
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Setting up a standard CYLINDRICAL, or POLAR, SPACE allows us to define the position of a particle at any point of its path using position vectors. As we did in the THREE DIMENSIONAL CARTESIAN SPACE, we have a vertical ZAXIS and replace the PLANAR X and Y-AXIS with our RADIAL and TRANSVERSE POSITION measurements, π and π: The NCEES doesnβt go in to using the THIRD DIMENSION in a CYCLINDRICAL SPACE, so we will forgoe discussion on it as well to avoid any confusion and further time in reviewing the material.
POSITION: The POSITION of any particle, is represented by a POSITION VECTOR from the origin to the location of this PARTICLE which can be expressed as: ππ = ππB NOTE, the NCEES does not clearly delineate the difference between r and r as it is presented in the Reference Handbook, therefore, we have chosen to denote one as ππ which represents the POSITION VECTOR of the PARTICLE whereas, r, is the RADIAL MEASUREMENT along the RADIAL AXIS.
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VELOCITY: In a POLAR COORDINATE SYSTEM, the VELOCITY is still the TIME DERIVATIVE of the POSITION VECTOR, such that:
π£=
πππ ππ‘
It has TWO COMPONENTS along the RADIAL and TRANSVERSE directions, presented in the NCEES Reference Handbook as: π£ = πππ + ππππ½ Where: π = π£B = TIME DERIVATIVE of the POSITION VECTOR ππ = π£E = POSITION VECTOR times the ANGULAR VELOCITY (π), which is the TIME derivative of the TRANSVERSE COMPONENT, π. Knowing these definitions allows us to write an equivalent expression as: π£ = π£B ππ + π£E ππ½
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ACCELERATION: As it is in the RECTANGULAR COORDINATE SYSTEM, the ACCELERATION is defined as the RATE OF CHANGE in the VELOCITY of a PARTICLE in motion, or in other terms, the TIME derivative of VELOCITY:
π=
ππ£ ππ‘
Or the SECOND TIME DERIVATIVE of the POSITION VECTOR: π4 π π= 4 ππ‘ It has TWO COMPONENTS along the RADIAL and TRANSVERSE directions, presented in the NCEES Reference Handbook as: π£ = (πβππ 4 )ππ + (ππ + 2ππ)ππ½ Where: (πβππ 4 ) = πB = π and is the second TIME derivative of the POSITION VECTOR ππ + 2ππ = πE = π and is the second TIME derivative of the TRANSVERSE COMPONENT, π, and is also known as the ANGULAR ACCELERATION Knowing these definitions allows us to write an equivalent expression as: π = πB ππ + πE ππ½
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LOCATING A POINT USING POLAR COORDINATES: Given a POINT, P, as illustrated:
To identify the POINT P in space, as illustrated in our rectangular coordinate system, we start from the origin (0,0) and go over to the right 2 UNITS and up 2 UNITS. This gives us the RECTANGULAR COORDINATES of POINT P as (2, 2). POLAR COORDINATES represents another way to locate a point in CYLINDRICAL COORDINATE PLANE. Using the same POINT P illustrated, we could start from the HORIZONTAL AXIS and rotate around the ORIGIN in a COUNTERCLOCKWISE DIRECTION a certain ANGLE of π. By CONVENTION, the HORIZONTAL POSITIVE X-AXIS and a ROTATION in a COUNTERCLOCKWISE DIRECTION constitute the starting point and positive rotation when working within a CYLINDRICAL COORDINATE SYSTEM.
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If we were to ROTATE CLOCKWISE, starting from the HORIZONTAL POSITIVE XAXIS, then we would get a NEGATIVE value for π. As we ROTATE around the ORIGIN, an IMAGINARY LINE some DISTANCE r from the ORIGIN will hit our POINT at P. This tells us that by ROTATING THROUGH an ANGLE ΞΈ and TRAVELING A DISTANCE r will also plot the POINT P that we have defined. The pair of POLAR COORDINATES can then be stated as (π, π), representing the POINT P in a CYLINDRICAL COORDINATE SYSTEM. NOTE that in POLAR COORDINATES, the ORIGIN, as we have referred to it as, is called the POLE and the X-AXIS is called the POLAR AXIS. Illustrating our POINT P in a POLAR COORDINATE SYSTEM, we have:
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CONVERTING FROM RECTANGULAR TO POLAR COORDINATES: To take any POINT P from a RECTANGULAR COORDINATE SYSTEM and QUANTIFY it in a CYLINDRICAL form, we will need to use a little knowledge from TRIGONOMETRY, and specifically, SOH CAH TOA and the PYTHAGOREAN THEOROM. Recall that SOH CAH TOA is a simple mnemonic that we learned back in our early days of MATHEMATICS to remember the various TRIGONMETRIC RELATIONSHIPS developed using RIGHT TRIANGLE geometry. These TRIGONMETRIC RELATIONSHIPS are defined as: π π¦ = β π π π₯ cos π = = β π π π¦ tan π = = π π₯ sin π =
The PYTHAGOREAN THEOREM also states: π₯4 + π¦4 = π4 For a more in depth review on using the various RELATIONSHIPS derived for SPECIAL CASE TRIANGLES, reference the MATHEMATICS section.
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Other relevant information regarding TRIGONOMETRY can also be referenced under the SUBJECT of MATHEMATICS on page 23 and 24 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Letβs revisit how we started this section with the RECTANGULAR COORDINATE (2, 2) plotted in a simple TWO DIMENSIONAL RECTANGULAR SPACE:
Letβs put some definition around our illustrated POINT, P, in the context of a POLAR COORDINATE SYSTEM. As stated, since we are going HORIZONTALLY 2 UNITS and up VERTICALLY by 2 UNITS, the RECTANGULAR COMPONENTS of POINT P can be defined as: π₯=2 π¦=2 However, we want to define this same POINT in the context of a POLAR COORDINATE SYSTEM.
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We do this by first recalling the convention of measurement in a POLAR COORDINATE SYSTEM, and second, applying our knowledge of TRIGONOMETRY that we outlined above. Recall that in POLAR COORDINATES, the ORIGIN is called the POLE and the POSITIVE X-AXIS is called the POLAR AXIS. By convention, the POLAR AXIS and a ROTATION in a COUNTERCLOCKWISE DIRECTION constitutes a STARTING POINT and the POSITIVE ROTATION. The pair of POLAR COORDINATES will be stated as (π, π), representing the POINT P in a POLAR, or CYLINDRICAL COORDINATE SYSTEM. The DISTANCE from the ORIGIN to the POINT, or otherwise, the r COORDINATE, is often referred to as the RADIAL POSITION COORDINATE. The ANGLE of ROTATION, π, is referred to as the TRANSVERSE COORDINATE. To begin the CONVERSION from RECTANGULAR COORDINATES to POLAR COORDINATES, we will being with the PYHTAGOREAN THEOREM, which again is stated as:: π₯4 + π¦4 = π4
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We have the X and Y RECTANGULAR COORDINATES, which allows us to get the RADIAL POSITION COORDINATE, r, or the LENGTH of the imaginary line from the ORIGIN to the POINT as: (2)4 + (2)4 = π 4 Rearranging and solving for the RADIAL POSITION COORDINATE, r, we get:
π=
2
4
+ 2
4
Or: π=2 2 Thatβs the first piece of our POLAR COORDINATE. Now we need to determine the TRANSVERSE COORDINATE, π, by using our knowledge of the TRIGONOMETRIC IDENTITIES and our trusted mnemonic SOH CAH TOA, which defines for us: π π¦ = β π π π₯ cos π = = β π π π¦ tan π = = π π₯ sin π =
With all of the x, y, and r COMPONENTS defined, we can use any of these relationships to determine the TRANSVERSE COORDINATE, π.
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Letβs use the SINE relationship:
sin π =
2 2 2
Rearranging and solving for π, we get:
π = sinQ3
2 2 2
Or: π = 45Β° With the RADIAL POSITION COORDINATE, r, and the TRANSVERSE COORDINATE, π, defined as: π=2 2 π = 45Β° We can now express the POINT P in POLAR FORM as: (2 2, 45Β°) It is important to NOTE that in the case of POLAR COORDINATES, the POINT P can also be expressed equivalently as having the COORDINATES: (2 2, 405Β°)
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π (2 2, ) 4 π (2 2, 2π + ) 4 And so on. All we are doing here is going from DEGREE to RADIANS and ADDING a FULL ROTATION, either as 360Β° or 2π.
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PARTICLE CURVILINEAR MOTION | CONCEPT EXAMPLE
The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
PROBLEM: At any instant, the horizontal position of the weather balloon can be mapped in space by the function, π₯ π‘ = 8π‘ ft, where t is given in seconds. If the path of flight can be represented by the formula π¦ =
3 3Z
π₯ 4 , the velocity of the balloon after 2 seconds in
flight is most close to: (ft/s)
A. 8 B. 26.8 C. 33.6 D. 64.2
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SOLUTION: The TOPIC of PARTICLE CURVILINEAR MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are given a scenario where a weather balloon is released in to the atmosphere, rising at a certain rate with a position that can be mapped using a pair of formulas, as defined in the problem statement: π₯ π‘ = 8t ft π¦=
1 4 π₯ 10
We are asked to determine the VELOCITY of this weather balloon 2 seconds in to its flight. The first step in analyzing any PARTICLE undergoing CURVILINEAR MOTION is setting up context around the motion using a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE. In this problem, we will be working with RECTANGULAR COORDINATES, therefore, we will operate within a standard CARTESIAN COORDINATE SYSTEM with the ORIGIN located at POINT A. This will give us a means to define the POSITION of our PARTICLE at a particular point in time using POSITION VECTORS.
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This POSITION VECTOR represents the TWO COMPONENTS, or DIMENSIONS, of the PARTICLE, expressed in VECTOR NOTATION as: π = π₯π + π¦π The CURVED PATH of the weather balloon flight represents a SEQUENCE of POSITIONS for the PARTICLE and is a FUNCTION OF TIME. Each COMPONENT is also operating as a FUNCTION OF TIME, as we see from the problem statement, which allows us to define where in space precisely it resides as it relates to a particular axis in our rectangular frame of reference. The POSITION COMPONENTS can be expressed as: π₯ = 8π‘ ft π¦=
1 (8π‘)4 10
This allows us to define the POSITION of the weather balloon at any point with the POSITION VECTOR:
π = 8t π +
1 (8t)4 π 10
Through our studies in CURVILINEAR MOTION, we know that the VELOCITY is the first TIME DERIVATIVE of POSITION, and can be represented in VECTOR TERMS as: π£ = π₯π + π¦π
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Remember that when we are cruising through the NCEES Reference Handbook, that the dots that fall above the x and y variables are specifically identifying those variables as TIME DERIVATIVES, or otherwise: ππ₯ = π£: ππ‘ ππ¦ π¦= = π£; ππ‘ π₯=
With this, we can define each TIME DERIVATIVE as: π(8t) ππ‘ 1 π 8t π¦ = 10 ππ‘ π₯=
4
Or: π₯=8 π¦=
128π‘ 10
This allows us to define the VELOCITY VECTOR as:
π£ = 8π +
128π‘ π 10
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At a TIME of 2 SECONDS, we get: π£ = 8π + 25.6π This is the VELOCITY, in VECTOR NOTATION, made up of TWO COMPONENTS, i and j, which are PERPENDICULAR to one another. With this, we can express the overall MAGNITUDE of the balloons VELOCITY at this point in TIME as:
π£=
(8)4 + (25.6)4
Or: π£ = 26.8 ft/s The MAGNITUDE of the VELOCITY of the weather balloon 2 seconds after it is released in to the atmosphere is 26.8 ft/s. The correct answer choice is B. ππ. π ππ/π¬
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