XX Relative Motion Concept Practice
RELATIVE MOTION | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
PROBLEM: Object A is traveling along a circular path with a velocity of 5 m/s that is increasing at a rate of 12 m/s 2 , as illustrated. If object B is traveling along a straight path with a speed that is decreasing at a rate of 5 m/s 2 , the relative velocity of object A is best represented as: m/s
A. 34 B. 84 + 56 C. 54 + 86 D. 504 + 126
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SOLUTION:
The TOPIC of RELATIVE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 and 73 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RELATIVE MOTION is the calculation of the MOTION of an OBJECT WITH REGARD TO SOME OTHER MOVING OBJECT, or stationary object for that matter. The motion is not calculated with reference to the earth, but the velocity of the object in reference to the other moving object as if it were in a static state. RELATIVE MOTION is a concept, and its calculation occurs in the context of RELATIVE VELOCITY, RELATIVE SPEED, or RELATIVE ACCELERATION. In our problem, we are given a pair of objects that are traveling along two independent tracks. Each object is defined using a combination of GEOMETRY, VELOCITY, and ACCELERATION:
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We have TWO MOVING OBJECTS, which means we will be dealing with a TRANSLATING FRAME moving within a FIXED FRAME, which will be the PLANE of the page. Since it is the RELATIVE VELOCITY of OBJECT A that we are concerned with in this problem, the TRANSLATING FRAME will sit on top of OBJECT B, such that:
Based on the problem statement, and the illustration given, we have the following data defined: Object A (Purple): (" = 5 m/s ," = 12 m/s 2 Object B (Orange): (# = 8 m/s ,# = −5 m/s 2
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These measurements are all RELATIVE to our FIXED FRAME, hence the usage of the VARIABLES (" , (# …etc. Again, we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. In VARIABLE terms, we are looking to define ("/# . Flipping over to PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and specifically, the SECTION titled RELATIVE MOTION, we can pull the formulas that we will be using to derive this measurement. The RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B can be found using: (" = (# + ("/# It is IMPORTANT to NOTE before moving forward in our computations, that this expression is given in VECTOR FORM and only holds true when we are dealing with a FIXED and TRANSLATING FRAME OF REFERENCE of PARTICLES. We are working with TRANSLATING FRAMES OF REFERENCE, so we are good there, but we don’t have our measurements in VECTOR FORM at this point, so let’s convert them in to CARTESIAN FORM. For OBJECT A, the VELOCITY is headed in the POSITIVE Y-DIRECTION, or rather: (" = 56 m/s
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We are told that the ACCELERATION at this point is INCREASING, or that it will be in the same DIRECTION as the VELOCITY. However, for an OBJECT traveling along the path of a CURVE, the ACCLERATION will be broken up in to TWO COMPONENTS, the NORMAL and TANGENTIAL ACCELERATION. At this point in the OBJECTS travel, or any point for that matter along a curved path, we can define a set of AXES coming directly from the OBJECT itself. These AXES are referred to as the TANGENTIAL AXIS (T-AXIS), which is TANGENT to the ARC, and the NORMAL AXIS (N-AXIS), which is pointing towards the CENTER OF CURVATURE and is NORMAL to the ARC. To illustrate this, generally we will have:
At any given time, we can set up a pair of AXIS, the TANGENTIAL and NORMAL AXIS, at the POSITION of this PARTICLE.
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It will always hold true that the TANGENTIAL AXIS will run TANGENT to the CURVE at the POSITION of the OBJECT and always points in the DIRECTION of the MOTION. The NORMAL AXIS will run NORMAL to the CURVE at the POSITION of the OBJECT and always points inward towards the CENTER OF CURVATURE.
In formulaic terms, the TANGENTIAL ACCLERATION is expressed as:
,8 = ( =
9( 9:
The NORMAL ACCLERATION is expressed as: (8 2 ,; = < Where: (8 = MAGNITUDE of VELOCITY at this POSITION < = INSTANTANEOUS RADIUS OF CURVATURE
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With the TANGENTIAL ACCELERATION and the NORMAL ACCELERATION components PERPENDICULAR to one another, we can express the MAGNITUDE of the OBJECT’S ACCELERATION at this POSITION as:
,=
,8 2 + ,; 2
So with all that being stated, expressing the ACCELERATION in terms of the TANGENTIAL and NORMAL COMPONENTS, we get: ,8 = 12 m/s 2 And: 5 m/s ,; = .5
2
Or: ,; = 50 m/s 2 We need to also express these values in CARTESIAN VECTOR FORM, which gives us: ,8 = 126 m/s 2 And: ,; = 104 m/s 2
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The ACCELERATION at any point on a curve will be expressed in terms of its NORMAL and TANGENTIAL COMPONENTS, such as for our case: , = 504 + 126 m/s 2 OBJECT B renders a much cleaner and quicker route to defining its VELOCITY and ACCELERATION in CARTESIAN VECTOR form since we aren’t dealing with any CURVES, just STRAIGHT LINE MOTION. However, the DIRECTIONS of each COMPONENT will be important to NOTE and if applied incorrectly, will be catastrophic to the conclusion of your computations. For OBJECT B, we know that it has some VELOCITY going to the LEFT, therefore: (# = −84 m/s We also know that OBJECT B is DECCELERATING at a rate of 5 m/s 2 , or: ,# = 54 m/s 2 NOTE that the HORIZONTAL COMPONENT for ACCELERATION is denoted as a POSITIVE VALUE. This is because of the fact that the OBJECT is DECCELERATING, or in other words, that the RATE OF SPEED is decreasing in an OPPOSITE DIRECTION of the MOTION of the OBJECT. Since the VELOCITY of our OBJECT in going to the LEFT, our ACCELERATION will be going in the opposite direction, to the RIGHT…or the POSITIVE X (j) DIRECTION,
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Alright, so with all that being stated, we have the following, all in CARTESIAN VECTOR FORM: OBJECT A: (" = 56 m/s ," = 504 + 126 m/s 2 OBJECT B: (# = −84 m/s ,# = 54 m/s 2 Recall that we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. The GENERAL FORMULA presented to us in the NCEES Reference Handbook for RELATIVE VELOCITY is: (" = (# + ("/# Plugging in the VECTORS we have set up, we get: 56 m/s = −84 m/s + ("/#
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Rearranging to solve for the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B, we have: ("/# = 84 m/s + 56 m/s Or in more friendly terms: ("/# = 84 + 56 m/s The correct answer choice is B. >? + @A B/C
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PROBLEM: Object A is traveling along a circular path with a velocity of 5 m/s that is increasing at a rate of 12 m/s 2 , as illustrated. If object B is traveling along a straight path with a speed that is decreasing at a rate of 5 m/s 2 , the relative velocity of object A is best represented as: m/s
A. 34 B. 84 + 56 C. 54 + 86 D. 504 + 126
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by Prepineer | Prepineer.com
SOLUTION:
The TOPIC of RELATIVE MOTION can be referenced under the main SUBJECT of DYNAMICS, and more specifically in the section titled PARTICLE KINEMATICS, on PAGE 72 and 73 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. RELATIVE MOTION is the calculation of the MOTION of an OBJECT WITH REGARD TO SOME OTHER MOVING OBJECT, or stationary object for that matter. The motion is not calculated with reference to the earth, but the velocity of the object in reference to the other moving object as if it were in a static state. RELATIVE MOTION is a concept, and its calculation occurs in the context of RELATIVE VELOCITY, RELATIVE SPEED, or RELATIVE ACCELERATION. In our problem, we are given a pair of objects that are traveling along two independent tracks. Each object is defined using a combination of GEOMETRY, VELOCITY, and ACCELERATION:
Made with
by Prepineer | Prepineer.com
We have TWO MOVING OBJECTS, which means we will be dealing with a TRANSLATING FRAME moving within a FIXED FRAME, which will be the PLANE of the page. Since it is the RELATIVE VELOCITY of OBJECT A that we are concerned with in this problem, the TRANSLATING FRAME will sit on top of OBJECT B, such that:
Based on the problem statement, and the illustration given, we have the following data defined: Object A (Purple): (" = 5 m/s ," = 12 m/s 2 Object B (Orange): (# = 8 m/s ,# = −5 m/s 2
Made with
by Prepineer | Prepineer.com
These measurements are all RELATIVE to our FIXED FRAME, hence the usage of the VARIABLES (" , (# …etc. Again, we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. In VARIABLE terms, we are looking to define ("/# . Flipping over to PAGE 72 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, and specifically, the SECTION titled RELATIVE MOTION, we can pull the formulas that we will be using to derive this measurement. The RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B can be found using: (" = (# + ("/# It is IMPORTANT to NOTE before moving forward in our computations, that this expression is given in VECTOR FORM and only holds true when we are dealing with a FIXED and TRANSLATING FRAME OF REFERENCE of PARTICLES. We are working with TRANSLATING FRAMES OF REFERENCE, so we are good there, but we don’t have our measurements in VECTOR FORM at this point, so let’s convert them in to CARTESIAN FORM. For OBJECT A, the VELOCITY is headed in the POSITIVE Y-DIRECTION, or rather: (" = 56 m/s
Made with
by Prepineer | Prepineer.com
We are told that the ACCELERATION at this point is INCREASING, or that it will be in the same DIRECTION as the VELOCITY. However, for an OBJECT traveling along the path of a CURVE, the ACCLERATION will be broken up in to TWO COMPONENTS, the NORMAL and TANGENTIAL ACCELERATION. At this point in the OBJECTS travel, or any point for that matter along a curved path, we can define a set of AXES coming directly from the OBJECT itself. These AXES are referred to as the TANGENTIAL AXIS (T-AXIS), which is TANGENT to the ARC, and the NORMAL AXIS (N-AXIS), which is pointing towards the CENTER OF CURVATURE and is NORMAL to the ARC. To illustrate this, generally we will have:
At any given time, we can set up a pair of AXIS, the TANGENTIAL and NORMAL AXIS, at the POSITION of this PARTICLE.
Made with
by Prepineer | Prepineer.com
It will always hold true that the TANGENTIAL AXIS will run TANGENT to the CURVE at the POSITION of the OBJECT and always points in the DIRECTION of the MOTION. The NORMAL AXIS will run NORMAL to the CURVE at the POSITION of the OBJECT and always points inward towards the CENTER OF CURVATURE.
In formulaic terms, the TANGENTIAL ACCLERATION is expressed as:
,8 = ( =
9( 9:
The NORMAL ACCLERATION is expressed as: (8 2 ,; = < Where: (8 = MAGNITUDE of VELOCITY at this POSITION < = INSTANTANEOUS RADIUS OF CURVATURE
Made with
by Prepineer | Prepineer.com
With the TANGENTIAL ACCELERATION and the NORMAL ACCELERATION components PERPENDICULAR to one another, we can express the MAGNITUDE of the OBJECT’S ACCELERATION at this POSITION as:
,=
,8 2 + ,; 2
So with all that being stated, expressing the ACCELERATION in terms of the TANGENTIAL and NORMAL COMPONENTS, we get: ,8 = 12 m/s 2 And: 5 m/s ,; = .5
2
Or: ,; = 50 m/s 2 We need to also express these values in CARTESIAN VECTOR FORM, which gives us: ,8 = 126 m/s 2 And: ,; = 104 m/s 2
Made with
by Prepineer | Prepineer.com
The ACCELERATION at any point on a curve will be expressed in terms of its NORMAL and TANGENTIAL COMPONENTS, such as for our case: , = 504 + 126 m/s 2 OBJECT B renders a much cleaner and quicker route to defining its VELOCITY and ACCELERATION in CARTESIAN VECTOR form since we aren’t dealing with any CURVES, just STRAIGHT LINE MOTION. However, the DIRECTIONS of each COMPONENT will be important to NOTE and if applied incorrectly, will be catastrophic to the conclusion of your computations. For OBJECT B, we know that it has some VELOCITY going to the LEFT, therefore: (# = −84 m/s We also know that OBJECT B is DECCELERATING at a rate of 5 m/s 2 , or: ,# = 54 m/s 2 NOTE that the HORIZONTAL COMPONENT for ACCELERATION is denoted as a POSITIVE VALUE. This is because of the fact that the OBJECT is DECCELERATING, or in other words, that the RATE OF SPEED is decreasing in an OPPOSITE DIRECTION of the MOTION of the OBJECT. Since the VELOCITY of our OBJECT in going to the LEFT, our ACCELERATION will be going in the opposite direction, to the RIGHT…or the POSITIVE X (j) DIRECTION,
Made with
by Prepineer | Prepineer.com
Alright, so with all that being stated, we have the following, all in CARTESIAN VECTOR FORM: OBJECT A: (" = 56 m/s ," = 504 + 126 m/s 2 OBJECT B: (# = −84 m/s ,# = 54 m/s 2 Recall that we are being asked to determine the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B. The GENERAL FORMULA presented to us in the NCEES Reference Handbook for RELATIVE VELOCITY is: (" = (# + ("/# Plugging in the VECTORS we have set up, we get: 56 m/s = −84 m/s + ("/#
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Rearranging to solve for the RELATIVE VELOCITY of OBJECT A as it relates to OBJECT B, we have: ("/# = 84 m/s + 56 m/s Or in more friendly terms: ("/# = 84 + 56 m/s The correct answer choice is B. >? + @A B/C
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