06 Curvilinear Motion Problem Set

PARTICLE CURVILINEAR MOTION | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: The position of a particle can be mapped in a rectangular space by the functions 𝑥 = 1 − 5𝑡 ' and 𝑦 = 2𝑡 * , where t is given in the units of seconds and the position measurement is given in meters. If the particle starts from a position of 𝑠 = 0 the magnitude of the particles total velocity after 3 seconds is closest to: m/s A. 4 B. 139 C. 16 D. 135.5



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PROBLEM 2: The position of a particle is defined by the function 𝑦 = 50 − .01𝑥 * , where x and y are given in the units of meters. If the particle starts from a position of 𝑠 = 0 and has a constant horizontal velocity of 3 m/s, the magnitude of the particles total velocity after 2 seconds is closest to: m/s A. . 36 B. 3.02 C. 9.4 D. 49.9



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PROBLEM 3: The motion of a piece of luggage dropping down a delivery chute is defined by the position vector 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m, where t is in seconds and the arguments for the sine and cosine are in radians. Disregarding all external forces, the location of the piece of luggage .75 seconds after it is dropped down the chute is closest to: A. . 15 m B. . 52 m C. . 68 m D. 1.03 m



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PROBLEM 4: The motion of a piece of luggage dropping down a delivery chute is defined by the position vector 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m, where t is in seconds and the arguments for the sine and cosine are in radians. Disregarding all external forces, the magnitude of the velocity of the piece of luggage .75 seconds after it is dropped down the chute is closest to: m/s A. 1.02 B. . 2 C. 2 D. . 5



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PROBLEM 5: The motion of a piece of luggage dropping down a delivery chute is defined by the position vector 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m, where t is in seconds and the arguments for the sine and cosine are in radians. Disregarding all external forces, the magnitude of the acceleration of the piece of luggage .75 seconds after it is dropped down the chute is closest to: m/s * A. 2.45 B. 4 C. . 2 D. 1.99



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PROBLEM 6: A 1 kg ball is being pushed by a rod along a vertical path where its position can be defined by the equation 𝑟 = . 5𝜃 m, with 𝜃 measured in radians. At the instant shown, if the ball has an angular velocity of 𝜃 = 2 rad/s and an angular acceleration of 𝜃 = 4 rad/s * , the magnitude of the acceleration is closest to: m/s *

A. 1.14 B. 6 C. 7.14 D. 7.23



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PARTICLE CURVILINEAR MOTION | PRACTICE PROBLEMS

SOLUTION 1: 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 PARTICLE undergoing CURVILINEAR MOTION with a position that can be mapped in time using a pair of formulas, as defined in the problem statement as: 𝑥 = 1 − 5𝑡 ' 𝑦 = 2𝑡 * We are also told that the PARTICLE starts at the POSITION: 𝑠=0 We are asked to determine the VELOCITY of this PARTICLE 3 seconds in to its defined MOTION. The first step in analyzing any PARTICLE undergoing CURVILINEAR MOTION is to set up context around the motion using a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE.

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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 (0,0). This will allow us to define the POSITION of our PARTICLE at a particular point in time using POSITION VECTORS. This POSITION VECTOR represents the TWO COMPONENTS, or DIMENSIONS, of the PARTICLE, expressed in VECTOR NOTATION as: 𝑟 = 𝑥𝒊 + 𝑦𝒋 The POSITION COMPONENTS at any point are defined for us in the problem statement as: 𝑥 = 1 − 5𝑡 ' 𝑦 = 2𝑡 * With this, we can define the POSITION VECTOR of the PARTICLE at any point as: 𝑟 = 1 − 5𝑡 ' 𝒊 + 2𝑡 * 𝒋 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: 𝑣 = 𝑥𝒊 + 𝑦𝒋



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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: 𝑑𝑥 = 𝑣J 𝑑𝑡 𝑑𝑦 𝑦= = 𝑣K 𝑑𝑡 𝑥=

With this, we can define each TIME DERIVATIVE as: 𝑑𝑥 = 15𝑡 * 𝑑𝑡 𝑑𝑦 𝑦= = 4𝑡 𝑑𝑡 𝑥=

This gives us the VELOCITY VECTOR defined as: 𝑣 = (15𝑡 * )𝒊 + (4𝑡)𝒋 At a TIME of 3 SECONDS, we get: 𝑣 = 15 3

*

𝒊 + 4(3) 𝒋

Or: 𝑣 = 135 𝒊 + 12 𝒋



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This is the VELOCITY, in VECTOR NOTATION, made up of TWO COMPONENTS, i and j, which are PERPENDICULAR to one another. We can express the overall MAGNITUDE of the particle’s VELOCITY at this point in TIME as:

𝑣=

(135)* + (12)*

Or: 𝑣 = 135.5 m/s The MAGNITUDE of the VELOCITY of this PARTICLE at a TIME of 3 seconds is 135.5 m/s. The correct answer choice is D. 𝟏𝟑𝟓. 𝟓 𝐦/𝐬

SOLUTION 2: 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 information about a PARTICLE undergoing CURVILINEAR MOTION with its POSITION in a TWO DIMENSIONAL space defined by the formula: 𝑦 = 50 − .01𝑥 *

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We are also told that the particle has the following characteristics, as defined: 𝑣J = 3 m/s 𝑠=0 We are asked to determine the VELOCITY of this particle 2 seconds in to its defined MOTION. The first step in analyzing any PARTICLE undergoing CURVILINEAR MOTION is to set 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 an ORIGIN located at (0,0). This allows us to define the POSITION of our PARTICLE at a particular point in time using POSITION VECTORS. This POSITION VECTOR represents the TWO COMPONENTS, or DIMENSIONS, of the PARTICLE, expressed in VECTOR NOTATION as: 𝑟 = 𝑥𝒊 + 𝑦𝒋 The CURVED PATH of MOTION represents a SEQUENCE of POSITIONS for the PARTICLE and is a FUNCTION OF TIME.



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Each COMPONENT is also operating as a FUNCTION OF TIME, which as we see in the problem statement, is given to us using a single function. This allows us to define where in space precisely this PARTICLE resides as it relates to a particular axis in our rectangular frame of reference. Let’s put in some work defining these POSITION COMPONENTS. We are told the the PARTICLE has a CONSTANT HORIZONTAL VELOCITY, or that: 𝑣J = 3 m/s Recall that VELOCITY is the FIRST TIME DERIVATIVE of VELOCITY, therefore:

𝑥=

𝑑𝑥 = 3 m/s 𝑑𝑡

Through INTEGRATION, we can define the X-COMPONENT of our POSITION VECTOR as:

𝑑𝑥 =

3 𝑑𝑡

Or: 𝑥 = 3𝑡



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This allows us to take the formula we are presented in the problem statement and define the Y-COMPONENT of our POSITION VECTOR as: 𝑦 = 50 − .01𝑥 * Plugging in the X-COMPONENT, as defined: 𝑦 = 50 − .01(3𝑡)* With each COMPONENT defined, the POSITION VECTOR of this PARTICLE at any point is: 𝑟 = 3t 𝒊 + 50 − .01(3𝑡)* 𝒋 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: 𝑣 = 𝑥𝒊 + 𝑦𝒋 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: 𝑑𝑥 = 𝑣J 𝑑𝑡 𝑑𝑦 𝑦= = 𝑣K 𝑑𝑡 𝑥=



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Recall that we are given the HORIZONTAL COMPONENT of the VELOCITY as:

𝑥=

𝑑𝑥 = 3 m/s 𝑑𝑡

We need to do some work to determine that VERTICAL COMPONENT of the VELOCITY. Recall that the PARTICLE is undergoing MOTION as defined by the function: 𝑦 = 50 − .01𝑥 * Since y is a function of x, we must use the CHAIN RULE to determine it’s DERIVATIVE as it relates to TIME, which conceptually is expressed as:

𝑦=

𝑑𝑦 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡

We have already established the latter part of this DERIVATION as: 𝑑𝑥 = 𝑣J = 3 m/s 𝑑𝑡 Which leaves us with the first part, which is: 𝑑𝑦 = −.02𝑥 𝑑𝑥



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Therefore: 𝑦 = (−.02𝑥)(3 m/s) = −.06𝑥 Or: 𝑦 = −.06𝑥 Plugging in the formula we have defined for the HORIZONTAL POSITION, X , we get: 𝑦 = −.06 3𝑡 = −.18𝑡 This gives us the VELOCITY VECTOR fully defined as: 𝑣 = 3 𝒊 + −.18𝑡 𝒋 At a TIME of 2 SECONDS, we get: 𝑣 = 3 𝒊 + −.36 𝒋 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 particle’s VELOCITY at this point in TIME as:

𝑣=



(3)* + (−.36)*

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Or: 𝑣 = 3.02 m/s The MAGNITUDE of the VELOCITY of the PARTICLE after 2 seconds of motion is 3.02 m/s. The correct answer choice is B. 𝟑. 𝟎𝟐 𝐦/𝐬

SOLUTION 3: 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 told that a piece of luggage is dropped down a delivery chute with its POSITION defined by the POSITION VECTOR: 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m We are asked to determine the LOCATION of this piece of luggage, relative to the ORIGIN, . 75 seconds after it is released down the chute. The first thing we typically do before analyzing any PARTICLE undergoing CURVILINEAR MOTION is aim to set up context around the motion using a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE. Made with by Prepineer | Prepineer.com

Fortunately, this problem does us a solid by defining the POSITION of our PARTICLE at any point in time using the POSITION VECTOR: 𝑟 = 0.5 sin 2𝑡 𝑖 + 0.5 cos 2𝑡 𝑗 − 0.2t 𝑘 m This piece of information, along with the TIME in which we are asked to analyze, gives us every we need to solve this problem. Plugging in the TIME value: 𝑡 = .75 𝑠 We get: 𝑟 = 0.5 sin 2(.75 𝑠) 𝒊 + 0.5 cos 2(.75 𝑠) 𝒋 − 0.2(.75 𝑠) 𝒌 m This gives us a POSITION VECTOR with the THREE COMPONENTS explicitly stated at this particular point in time as: 𝑟 = . 498𝒊 + .035𝒋 + .15𝒌 m The MAGNITUDE of this VECTOR represents the DISTANCE from the ORIGIN to the point of the luggage after .75 seconds, and can be expressed as:

𝑟=



(.498)* + (.035)* + (.15)*

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Or: 𝑟 = .52 m The LOCATION of the luggage .75 seconds after it is dropped down the chute is .52 meters from the ORIGIN. The correct answer choice is B. . 𝟓𝟐 𝐦

SOLUTION 4: 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 told that a piece of luggage is dropped down a delivery chute with its POSITION defined by the POSITION VECTOR: 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m We are asked to determine the MAGNITUDE of the VELOCITY of this piece of luggage . 75 seconds after it is released down the chute. 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. Made with by Prepineer | Prepineer.com

Fortunately, this problem defines the POSITION of our PARTICLE at a particular point in time using the POSITION VECTOR: 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m This piece of information, along with the TIME in which we are asked to analyze, is a great start in getting to our solution. Recall that VELOCITY is the first TIME DERIVATIVE of POSITION and is represented in VECTOR TERMS as: 𝑣 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 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 identifying those variables as TIME DERIVATIVES, or otherwise: 𝑑𝑥 = 𝑣J 𝑑𝑡 𝑑𝑦 𝑦= = 𝑣K 𝑑𝑡 𝑑𝑧 𝑧= = 𝑣Z 𝑑𝑡 𝑥=

This shows us that we can also write the VELOCITY vector in terms of its VELOCITY COMPONENTS, such that: 𝑣 = 𝑣J 𝒊 + 𝑣K 𝒋 + 𝑣Z 𝒌

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With this, we can define each TIME DERIVATIVE as:

𝑥=

𝑑 0.5 sin 2𝑡 𝑑𝑡

𝑑 0.5 cos 2𝑡 𝑑𝑡 𝑑 −0.2t 𝑧= 𝑑𝑡 𝑦=

Or: 𝑥 = cos 2𝑡 𝑦 = −sin 2𝑡 𝑧 = −.2 With this, the VELOCITY VECTOR can be expressed as: 𝑣 = cos 2𝑡 𝒊 − sin 2𝑡 𝒋 − .2𝒌 At a TIME of .75 SECONDS, we derive a MAGNITUDE of the VELOCITY as:

𝑣=

cos 2(.75 𝑠)

*

+ −sin 2(.75 𝑠)

*

+ (−.2)*

Or: 𝑣 = 1.02 m/s The MAGNITUDE of the VELOCITY of the luggage after .75 seconds is 1.02 m/s. Made with by Prepineer | Prepineer.com

The correct answer choice is A. 𝟏. 𝟎𝟐 𝐦/𝐬

SOLUTION 5: 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 told that a piece of luggage is dropped down a delivery chute with its POSITION defined by the POSITION VECTOR: 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m We are asked to determine the MAGNITUDE of the ACCELERATION. 75 seconds after this piece of luggage is released down the chute. The first thing we typically do before analyzing any PARTICLE undergoing CURVILINEAR MOTION is set up context around the motion using a STANDARD RECTANGULAR or CYLINDRICAL COORDINATE SPACE. Fortunately, this problem defines the POSITION of our PARTICLE at a particular point in time using the POSITION VECTOR: 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m



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This piece of information, along with the TIME in which we are asked to analyze, is a great start in getting to our solution. Recall that ACCELERATION is defined as the RATE OF CHANGE in the VELOCITY of a PARTICLE in motion. It may also be referred as the SECOND TIME DERIVATIVE of the POSITION: 𝑑* 𝑟 𝑎= * 𝑑𝑡 Expressed in VECTOR TERMS, we have: 𝑎 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 In this case, the double dots that fall above the x, y, and z variables are specifically identifying those variables as SECOND TIME DERIVATIVES of the POSITION COMPONENTS, or otherwise: 𝑑* 𝑥 𝑥 = * = 𝑎J 𝑑𝑡 𝑑* 𝑦 𝑦 = * = 𝑎K 𝑑𝑡 𝑑* 𝑧 𝑧 = * = 𝑎Z 𝑑𝑡



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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. Using the definitions as stated, we can also express the ACCELERATION VECTOR in terms of its ACCELERATION COMPONENTS as: 𝑎 = 𝑎J 𝒊 + 𝑎K 𝒋 + 𝑎Z 𝒌 Rallying back to the POSITION VECTOR as defined in the problem statement, we have: 𝑟 = 0.5 sin 2𝑡 𝒊 + 0.5 cos 2𝑡 𝒋 − 0.2t 𝒌 m With this, lets define the FIRST TIME DERIVATIVE of each COMPONENT: 𝑑 0.5 sin 2𝑡 𝑑𝑡 𝑑 0.5 cos 2𝑡 𝑦= 𝑑𝑡 𝑑 −0.2t 𝑧= 𝑑𝑡 𝑥=

Or: 𝑥 = cos 2𝑡 𝑦 = −sin 2𝑡 𝑧 = −.2

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Taking these newly defined COMPONENTS, lets now define the SECOND TIME DERIVATIVE, we have: 𝑑 cos 2𝑡 = −2 sin 2𝑡 𝑑𝑡 𝑑 −sin 2𝑡 𝑦= = −2 cos 2𝑡 𝑑𝑡 𝑑 −.2 𝑧= =0 𝑑𝑡 𝑥=

Or: 𝑥 = −2 sin 2𝑡 𝑦 = −2 cos 2𝑡 𝑧=0 Which gives us all the ACCELERATION COMPONENTS and the ACCELERATION VECTOR: 𝑎 = −2 sin 2𝑡 𝒊 − 2 cos 2𝑡 𝒋 + 0 𝒌 At a TIME of .75 SECONDS, we derive a MAGNITUDE of the ACCELERATION as:

𝑎=

−2 sin 2(.75 𝑠)

*

+ −2 cos 2(.75 𝑠)

*

+ (0)*

Or:



𝑎 = 1.99 m/s * Made with

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The MAGNITUDE of the ACCELERATION of the luggage after .75 seconds is 1.99 m/s * . The correct answer choice is D. 𝟏. 𝟗𝟗 𝐦/𝐬 𝟐

SOLUTION 6: 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 ball is being pushed up by a rod assembly, undergoing motion based on certain characteristics, as defined in the problem statement: 𝑟 = . 5𝜃 m 𝜃 = 2 rad/s 𝜃 = 4 rad/s * We are asked to determine the MAGNITUDE of the ball’s ACCELERATION when the POSITION of the assembly has been rotated an ANGLE of 𝜃 relative to the REFERENCE AXIS.



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This ANGLE is given to us as:

𝜃=

π 2

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. Although we are always fans of working in a RECTANGULAR SPACE, for simplicity sake, in this problem, we have been given everything in CYLINDRICAL COORDINATES…and with that, we will set up and operate within a standard CYLINDRICAL COORDINATE SYSTEM with the ORIGIN set at POINT O. The ACCELERATION of a PARTICLE undergoing CURVILINEAR MOTION is defined as the RATE OF CHANGE in the VELOCITY of a PARTICLE in motion, or in other terms, the TIME DERIVATIVE of VELOCITY:

𝑎=

𝑑𝑣 𝑑𝑡

Which can also be expressed as the SECOND TIME DERIVATIVE of the POSITION VECTOR: 𝑑* 𝑟 𝑎= * 𝑑𝑡



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The ACCELERATION will have TWO COMPONENTS, one each along the RADIAL and TRANSVERSE directions, and is presented in the NCEES Reference Handbook as: 𝑎 = (𝑟−𝑟𝜃 * )𝒆𝒓 + (𝑟𝜃 + 2𝑟𝜃)𝒆𝜽 Where: (𝑟−𝑟𝜃 * ) = 𝑎a = 𝑟 and is the second TIME derivative of the POSITION VECTOR 𝑟𝜃 + 2𝑟𝜃 = 𝑎b = 𝜃 and is the second TIME derivative of the TRANSVERSE COMPONENT, 𝜃…also referred to as the ANGULAR ACCELERATION These definitions allow us to write an equivalent expression for the ACCELERATION VECTOR as: 𝑎 = 𝑎a 𝒆𝒓 + 𝑎b 𝒆𝜽 Highlighting the TWO COMPONENTS we will need to define, we have: 𝑎a = (𝑟−𝑟𝜃 * ) 𝑎b = 𝑟𝜃 + 2𝑟𝜃 As we have already identified above, the problem statement gives us a bunch of this information right up front, specifically: 𝑟 = . 5𝜃 π 𝜃= 2

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𝜃 = 2 rad/s 𝜃 = 4 rad/s * We need to put definition around the FIRST and SECOND TIME DERIVATIVE of the POSITION VECTOR, 𝑟 and 𝑟, and we should have all that we need to get this problem dialed in. The FIRST TIME DERIVATIVE of the POSITION VECTOR can be expressed as:

𝑟=

𝑑𝑟 𝑑𝑡

Because the POSITION is given to us in terms of the TRANSVERSE ANGLE, and not TIME, we need to use the CHAIN RULE, which equivalently is expressed as:

𝑟=

𝑑𝑟 𝑑𝜃 ∙ 𝑑𝜃 𝑑𝑡

Which when worked out gives us: 𝑟 = . 5𝜃 The SECOND TIME DERIVATIVE of the POSITION VECTOR can be expressed as:

𝑟=



𝑑𝑟 𝑑𝑡

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Which when worked out gives us: 𝑟 = . 5𝜃 We now have the following DATA defined: 𝑟 = . 5𝜃 𝑟 = . 5𝜃 𝑟 = . 5𝜃 𝜃=

π 2

𝜃 = 2 rad/s 𝜃 = 4 rad/s * Referring back to the GENERAL FORMULA for the ACCELERATION, we have: 𝑎 = 𝑎a 𝒆𝒓 + 𝑎b 𝒆𝜽 With each individual COMPONENT defined as: 𝑎a = (𝑟−𝑟𝜃 * ) 𝑎b = 𝑟𝜃 + 2𝑟𝜃 Plugging in our values we get: π 𝑎a = .5(4 rad/s * ) − (.5( )(2 rad/s)* 2

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And: π 𝑎b = .5( )(4 rad/s * ) + 2 . 5(2 rad/s)(2 rad/s) 2 Or: 𝑎a = −1.14 m/s * 𝑎b = 7.14 m/s * Bringing these TWO COMPONENTS together, we can define the ACCELERATION at this point as: 𝑎 = (−1.14 m/s * )𝒆𝒓 + (7.14 m/s * )𝒆𝜽 Remember that this is the ACCELERATION, in VECTOR NOTATION, made up of the RADIAL and TRANSVERSE COMPONENTS, which are PERPENDICULAR to one another. With this, we can express the overall MAGNITUDE of the balls ACCELERATION at this point in TIME as:

𝑎=

(−1.14 m/s * )* + (7.14 m/s * )*

Or: 𝑎 = 7.23 m/s *

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by Prepineer | Prepineer.com

The MAGNITUDE of the ACCELERATION of the ball is 7.23 m/s * . The correct answer choice is D. 𝟕. 𝟐𝟑 𝐦/𝐬 𝟐



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