Project

XII.) Kinematics of a Particle • Where are we?

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Particle Kinematics

Rectilinear Motion

Curvilinear Motion

Single Coordinates

Rectangular Coordinates

Projectile Motion

Constant Acceleration?

Constant Acceleration Equations

Yes NO Differential Relationships

Normal and Tangential

Coordinates

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian.

• When the path of motion is known in the plane, normal and tangential coordinates are often used.

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– i.) General Setup • Suppose we have a particle moving in the plane as

shown:

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• The origin of the coordinate system is coincident with the particle at the instant considered

• At an instant in time, we define the tangential direction to the path as �𝑢𝑢𝑡𝑡

• The positive direction of �𝑢𝑢𝑡𝑡 is along the path such that 𝑠𝑠 is increasing

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Define a normal direction as perpendicular to the tangential direction, given by �𝑢𝑢𝑛𝑛

• Note that �𝑢𝑢𝑛𝑛 has a positive sense directed towards the center of curvature

• The relations between 𝑎𝑎𝑡𝑡, 𝑣𝑣, 𝑡𝑡, and 𝑠𝑠 are the same for rectilinear motion. That is,

𝑎𝑎𝑡𝑡 = �̇�𝑣 or 𝑎𝑎𝑡𝑡𝑑𝑑𝑠𝑠 = 𝑣𝑣𝑑𝑑𝑣𝑣

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– EX. • Set up a normal and tangential coordinate system

at different points along the path

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𝑠𝑠

𝑠𝑠

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– EX. • Set up a normal and tangential coordinate system

at different points along the path

7

𝑠𝑠 �𝑢𝑢𝑛𝑛 �𝑢𝑢𝑛𝑛

�𝑢𝑢𝑛𝑛�𝑢𝑢𝑡𝑡

�𝑢𝑢𝑡𝑡�𝑢𝑢𝑡𝑡

𝑠𝑠 �𝑢𝑢𝑛𝑛

�𝑢𝑢𝑡𝑡

�𝑢𝑢𝑛𝑛�𝑢𝑢𝑛𝑛

�𝑢𝑢𝑡𝑡

�𝑢𝑢𝑡𝑡

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• We can define a radius of curvature, 𝜌𝜌, where two normals intersect for a small 𝑑𝑑𝑠𝑠

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Radius of CurvatureB. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• We will need the radius of curvature for the acceleration equations

• If the path is expressed as 𝑦𝑦 = 𝑓𝑓 𝑥𝑥 , the radius of curvature (from calculus) is given by:

𝜌𝜌 = 1 + 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2 3/2

𝑑𝑑2𝑦𝑦 𝑑𝑑𝑥𝑥2

• Note that 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

and 𝑑𝑑 2𝑑𝑑

𝑑𝑑𝑑𝑑2 are not time derivatives

• For a circle, 𝜌𝜌 = 𝑟𝑟 (the radius of the circle)

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– ii.) Velocity • At a particular instant, the velocity is given by:

�⃗�𝑣 = 𝑣𝑣�𝑢𝑢𝑡𝑡 • a.) Magnitude

𝑣𝑣 = �⃗�𝑣 = 𝑑𝑑𝑠𝑠 𝑑𝑑𝑡𝑡

= �̇�𝑠 = 𝑣𝑣2

• b.) Direction �𝑢𝑢𝑡𝑡

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�𝑢𝑢𝑡𝑡 B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– iii.) Acceleration • We can differentiate the velocity vector. Thus,

�⃗�𝑎 = 𝑑𝑑�⃗�𝑣 𝑑𝑑𝑡𝑡

= 𝑑𝑑 𝑑𝑑𝑡𝑡

(𝑣𝑣�𝑢𝑢𝑡𝑡)

= 𝑑𝑑 𝑑𝑑𝑡𝑡 𝑣𝑣 �𝑢𝑢𝑡𝑡 + 𝑣𝑣

𝑑𝑑 𝑑𝑑𝑡𝑡 �𝑢𝑢𝑡𝑡

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

• a.) Magnitude

�⃗�𝑎 = 𝑎𝑎𝑡𝑡 2 + 𝑎𝑎𝑛𝑛2

= �̇�𝑣2 + 𝑣𝑣 2

𝜌𝜌

2

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– iii.) Acceleration • b.) Direction

– The direction is given by:

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– iv.) Acceleration Derivation • We can differentiate the velocity vector. Thus,

�⃗�𝑣 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣�̇𝑢𝑢𝑡𝑡

• We can say that: �𝑢𝑢𝑡𝑡′ = �𝑢𝑢𝑡𝑡 + 𝑑𝑑�𝑢𝑢𝑡𝑡

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? ? ?

�𝑢𝑢𝑡𝑡 𝑑𝑑�𝑢𝑢𝑡𝑡

�𝑢𝑢′𝑡𝑡 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 𝜌𝜌 𝜌𝜌

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• We see that the direction of �𝑢𝑢𝑡𝑡 is in the direction of �𝑢𝑢𝑛𝑛, but what is the magnitude?

𝑑𝑑�𝑢𝑢𝑡𝑡 = 𝑑𝑑𝑑𝑑 ∴ 𝑑𝑑�𝑢𝑢𝑡𝑡 = 𝑑𝑑𝑑𝑑�𝑢𝑢𝑛𝑛

�̇𝑢𝑢𝑡𝑡 = �̇�𝑑�𝑢𝑢𝑛𝑛 But we know that:

𝑑𝑑𝑠𝑠 = 𝜌𝜌𝑑𝑑𝑑𝑑 �̇�𝑠 = 𝜌𝜌�̇�𝑑

�̇�𝑑 = �̇�𝑠 𝜌𝜌

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• By substitution: �̇𝑢𝑢𝑡𝑡 = �̇�𝑑�𝑢𝑢𝑛𝑛

= �̇�𝑠 𝜌𝜌 �𝑢𝑢𝑛𝑛

= 𝑣𝑣 𝜌𝜌 �𝑢𝑢𝑛𝑛

• Putting this all together �⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣�̇𝑢𝑢𝑡𝑡

= �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣 𝑣𝑣 𝜌𝜌 �𝑢𝑢𝑛𝑛

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𝑎𝑎𝑡𝑡 𝑎𝑎𝑛𝑛

B. Slaboch 2017

XII.) Kinematics of a Particle • Cartesian vs. Normal and Tangential

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𝑟𝑟 = 𝑥𝑥 ̂𝚤𝚤 + 𝑦𝑦 ̂𝚥𝚥 + 𝑧𝑧�𝑘𝑘Position

Velocity

Acceleration

�⃗�𝑣 = �̇�𝑥 ̂𝚤𝚤 + �̇�𝑦 ̂𝚥𝚥 + �̇�𝑧�𝑘𝑘

�⃗�𝑎 = �̈�𝑥 ̂𝚤𝚤 + �̈�𝑦 ̂𝚥𝚥 + �̈�𝑧�𝑘𝑘

Rectangular Coordinates Normal and Tangential

Coordinates

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

𝑟𝑟 = 0

�⃗�𝑣 = 𝑣𝑣�𝑢𝑢𝑡𝑡

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– EX. • Determine the velocity and acceleration at the

instant shown. The particle has a constant speed of 2 m/s. What is the magnitude of the acceleration at this instant?

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Given: – The path, given by 𝑦𝑦 = 2𝑥𝑥2

– Constant speed, 𝑣𝑣 = 2 m/s, �̇�𝑣 = �̈�𝑥 = 0 m/s2

• Find: – �⃗�𝑣, �⃗�𝑎, �⃗�𝑎

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – The velocity in rectangular components is given by:

�⃗�𝑣 = �̇�𝑥 ̂𝚤𝚤 + �̇�𝑦 ̂𝚥𝚥 – From the plot, we can infer that �̇�𝑥 = −2 m

s

– From the given equations, we can calculate �̇�𝑦 as: 𝑦𝑦 = 2𝑥𝑥2 �̇�𝑦 = 4𝑥𝑥�̇�𝑥

= 4 0 −2 = 0 m

s

– We could also calculate �̇�𝑦 by inspection!

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – Thus,

�⃗�𝑣 = −2 ̂𝚤𝚤 + 0 ̂𝚥𝚥 and therefore,

�⃗�𝑣 = �̇�𝑥2 + �̇�𝑦2

= −2 2 + 02

= 2 m s

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – The acceleration in rectangular components is given by:

�⃗�𝑎 = �̈�𝑥 ̂𝚤𝚤 + �̈�𝑦 ̂𝚥𝚥 – we can calculate �̈�𝑦 as:

𝑦𝑦 = 2𝑥𝑥2 �̇�𝑦 = 4𝑥𝑥�̇�𝑥 �̈�𝑦 = 4(𝑥𝑥�̈�𝑥 + �̇�𝑥2)

Thus, �⃗�𝑎 = 0 ̂𝚤𝚤 + 4�̇�𝑥2 ̂𝚥𝚥 �⃗�𝑎 = 0 ̂𝚤𝚤 + 4(−2)2 ̂𝚥𝚥 �⃗�𝑎 = 16 ̂𝚥𝚥

The magnitude is then �⃗�𝑎 = 16 m s2

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– EX. Let’s rework the problem using N-T coordinates

• Determine the velocity and acceleration at the instant shown. The particle has a constant speed of 2 m/s. What is the magnitude of the acceleration at this instant?

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – Before doing anything with normal and tangential

components, we need a coordinate system!

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�𝑢𝑢𝑡𝑡

�𝑢𝑢𝑛𝑛

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – The velocity vector in normal and tangential components

is given by: �⃗�𝑣 = 𝑣𝑣�𝑢𝑢𝑡𝑡

– Substituting in the known value of 𝑣𝑣 gives: �⃗�𝑣 = 2�𝑢𝑢𝑡𝑡

– The magnitude is then: �⃗�𝑣 = 2

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – The acceleration vector in normal and tangential

components is given by:

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

– We know �̇�𝑣 and 𝑣𝑣 from the givens, but we need to find 𝜌𝜌

𝜌𝜌 = 1 + 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

2 3/2

𝑑𝑑2𝑦𝑦 𝑑𝑑𝑥𝑥2

= 1+ 4𝑑𝑑 2

3/2

4 = 1+0

3/2

4 = 1

4 m

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – Evaluating the individual components, we now have:

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

= 0�𝑢𝑢𝑡𝑡 + 22 1 4 �𝑢𝑢𝑛𝑛

= 16�𝑢𝑢𝑛𝑛 �⃗�𝑎 = 02 + 162

= 16 m s2

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

– EX. • Starting from rest, a motorboat travels around a

circular path of 𝜌𝜌 = 50 m at a speed that increases with time, 𝑣𝑣 = (0.2 𝑡𝑡2) m/s. What is the magnitude of the boat’s velocity and magnitude of the boat’s acceleration at the instant 𝑡𝑡 = 3 s.

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�⃗�𝑣

𝜌𝜌

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – We can start by adding a coordinate system

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�𝑢𝑢𝑡𝑡

�𝑢𝑢𝑛𝑛

�⃗�𝑣

𝜌𝜌

B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Given: – Speed the boat as a function of time – Radius of the circular path, 𝜌𝜌 = 50 m – The boat start time from rest, 𝑣𝑣 0 = 0

• Find: – Magnitude of the velocity at 𝑡𝑡 = 3 s, �⃗�𝑣(3) =? – Magnitude of the acceleration at 𝑡𝑡 = 3 s, �⃗�𝑎(3) =?

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – The velocity can be written as:

�⃗�𝑣 = 𝑣𝑣�𝑢𝑢𝑡𝑡 – The magnitude can be expressed as:

�⃗�𝑣 = 0.2𝑡𝑡2 = 0.2 3 2 = 1.8 m/s – The acceleration can be calculated as:

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

– Differentiating the speed gives: �̇�𝑣 = 0.4𝑡𝑡 �̇�𝑣 3 = 0.4 3 = 1.2 m

s2

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B. Slaboch 2017

XII.) Kinematics of a Particle • F.) Normal and Tangential Components

• Execution: – By substitution:

�⃗�𝑎 3 = 1.2�𝑢𝑢𝑡𝑡 + 1.8 2

50 �𝑢𝑢𝑛𝑛

= 1.2�𝑢𝑢𝑡𝑡 + 1.8 2

50 �𝑢𝑢𝑛𝑛

m s2

– We can then find the magnitude by:

�⃗�𝑎(3) = 1.2 2 + 1.8 2

50

2

= 1.2017 m s2

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B. Slaboch 2017

FE Exam Style Problems An object weighs 2 pounds and is travelling in a circular path of radius 5 ft. at a constant speed of 10 feet per second. The magnitude of the acceleration of the object is most nearly:

a.) 20 ft s2

b.) 16 ft s2

c.) 18 ft s2

d.) 17 ft s2

e.) 14 ft s2

FE Exam Style Problems An object weighs 2 pounds and is travelling in a circular path of radius 5 ft. at a constant speed of 10 feet per second. The magnitude of the acceleration of the object is most nearly:

a.) 𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟 𝐬𝐬𝟐𝟐

b.) 16 ft s2

c.) 18 ft s2

d.) 17 ft s2

e.) 14 ft s2

Normal-Tangential

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑡𝑡 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

�⃗�𝑎 = 0�𝑢𝑢𝑡𝑡 + 102

5 �𝑢𝑢𝑛𝑛

�⃗�𝑎 = 20�𝑢𝑢𝑛𝑛

�⃗�𝑎 = 20 ft s2