Rectangular Cylindrical Coordinates Lecture 8 MEC2023F Polar Coordinates Cylindrical Coordinates Cylindrical Coordinates

2017/03/29

Applications

MEC2023F

MEC2023F

A cylindrical coordinate system can be used in certain cases where the particle moves along a 3-D curve.

Rectangular Cylindrical Coordinates

In the figure shown, the box slides down the helical ramp. How would you find the box’s velocity components to check to see if the package will fly off the ramp?

Lecture 8

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Applications

MEC2023F

The cylindrical coordinate system can be used to describe the motion of the girl on this slide. Here the radial coordinate is constant, the transverse coordinate increases with time as the girl rotates about the vertical axis, and her altitude, z, decreases with time. How can you find her acceleration components ? © 2013 Pearson Education, Inc.

MEC2023F

In the previous section we derived equations for position, velocity and acceleration in polar notation, but this was limited to two dimensions. 𝑟 = 𝑟𝑒𝑟 = 𝑥𝑖 + 𝑦𝑗 𝑣 = 𝑟𝑒𝑟 + 𝑟 𝜃𝑒𝜃 𝑎 = 𝑟 − 𝑟𝜃 2 𝑒𝑟 + 𝑟𝜃 + 2𝑟𝜃 𝑒𝜃 © 2013 Pearson Education, Inc.

Cylindrical Coordinates

MEC2023F

If the particle P moves along a space curve, its position can be written as 𝑅 = 𝑟𝑒𝑟 + 𝑧𝑘

To obtain expressions for the velocity and acceleration, the derivatives of unit vectors 𝑒𝑟 , 𝑒𝜃 and 𝑘 in terms of the coordinates r, θ and z are required. By inspection of the figure above it can be seen that a change in r or z would not change the direction of the unit vectors. By contrast, a change in θ would alter the direction of 𝑒𝑟 and 𝑒𝜃 , but not 𝑘 . © 2013 Pearson Education, Inc.

Polar Coordinates

Cylindrical Coordinates Starting with the position vector

MEC2023F

𝑅 = 𝑟 𝑒𝑟 + 𝑧𝑘

The time derivative of position gives the velocity: 𝑣 = 𝑅 = 𝑟𝑒𝑟 + 𝑟𝑒𝑟 + 𝑧𝑘 𝑣 = 𝑟𝑒𝑟 + 𝑟 𝜃𝑒𝜃 + 𝑧𝑘 The second time derivative gives the acceleration: 𝑎 = 𝑣 = 𝑅 = 𝑟 − 𝑟𝜃 2 𝑒𝑟 + 𝑟 𝜃 + 2𝑟𝜃 𝑒𝜃 + 𝑧𝑘

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2017/03/29

Example 8.1

MEC2023F

Solution 8.1

MEC2023F

𝑟 = (3 + 1.5𝑡) m

𝜃 = (0.5𝑡) rad

𝑧 = 4𝑡 2 m

𝑟 = 1.5 m/s

𝜃 = 0.5 rad/s

𝑧 = (8𝑡) m/s

𝑟=0

𝜃=0

𝑧 = 8 m/s 2

At 𝑡 = 3 𝑠

The arm of the robot is extending at a constant rate 𝑟 = 1.5 m/s when r = 3 m, z = (4t2) m, and θ = (0.5t) rad, where t is in seconds. Find the velocity and acceleration of the grip A when t = 3 s.

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Example 8.2

𝑟 = 3m

𝜃 = 1.5 rad

𝑧 = 36 m

𝑟 = 1.5 m/s

𝜃 = 0.5 rad/s

𝑧 = 24 m/s

𝑟=0

𝜃=0

𝑧 = 8 m/s2

Velocity

Acceleration

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MEC2023F

𝑣 = 𝑟𝑒𝑟 + 𝑟𝜃 𝑒𝜃 + 𝑧𝑘 𝑣 = 1.5𝑒𝑟 + (3)(0.5)𝑒𝜃 + 24𝑘 𝑣 = 1.5𝑒𝑟 + 1.5𝑒𝜃 + 24𝑘

𝑎 = 𝑟 − 𝑟𝜃 2 𝑒𝑟 + 𝑟𝜃 + 2𝑟𝜃 𝑒𝜃 + 𝑧𝑘 𝑎 = −0.75𝑒𝑟 + 1.5𝑒𝜃 + 8𝑘 𝑎 = 8.17 m/s2

Solution 8.2 𝑟 = 0.5 m

The box slides down the helical ramp which is defined by 𝑟 = 0.5 m, 𝜃 = 0.5𝑡 3 rad and 𝑧 = 2 − 0.2𝑡 2 , where t is in seconds.

𝑧 =2−

m

𝑧 = −0.4𝑡 m/s

𝑟=0

𝜃 = 3.0𝑡

𝑧 = −0.4 m/s 2

𝑟 = 0.5 m

𝜃 = 6.28 rad

𝑧 = 0.919 m

𝑟=0

𝜃 = 8.11 rad/s

𝑧 = −0.93 m/s

𝑟=0

𝜃 = 6.98 rad/s2

𝑧 = −0.4 m/s2

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MEC2023F

𝑣 = 𝑟𝑒𝑟 + 𝑟𝜃 𝑒𝜃 + 𝑧𝑘 𝑣 = 0 + 4.05𝑒𝜃 − 0.93𝑘

𝑣 = 4.16 m/s

𝑎 = 𝑟 − 𝑟𝜃 2 𝑒𝑟 + 𝑟𝜃 + 2𝑟𝜃 𝑒𝜃 + 𝑧𝑘 𝑎 = −32.9𝑒𝑟 + 3.49𝑒𝜃 − 0.4𝑘 𝑎 = 33.1 m/s2

Solution 8.3

MEC2023F

𝑟 = 0.4 m

𝜃

𝑟=0

𝜃

𝑧 = −0.2𝜃 m/s

𝑟=0

𝜃

𝑧 = −0.2𝜃 m/s 2

𝑧 = −0.2𝜃 m

𝑟 = 0.5 m

𝜃

𝑧 = −0.2𝜃 m

𝑟=0

𝜃 = 6 rad/s

𝑧 = −1.2 m/s

𝑟=0

𝜃 = −10 rad/s2

𝑧 = 2.0 m/s2

At the instant given

Velocity

Acceleration

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rad

𝜃 = 1.5𝑡 2 rad/s

Velocity

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Determine the magnitudes of the velocity and acceleration of the box at this instant.

𝜃=

0.2𝑡 2

𝑟=0

Acceleration

A particle moves along a spiral described in cylindrical coordinates by 𝑅 = 0.4 m and 𝑧 = −0.2𝜃 m. At a certain instant 𝜃 = 6 𝑟𝑎𝑑/𝑠 and 𝜃 = −10 𝑟𝑎𝑑/𝑠 2,

MEC2023F 0.5𝑡 3

When 𝜃 = 2𝜋 rad then 𝑡 = 2.33 𝑠

Determine the magnitudes of the velocity and acceleration of the box at the instant 𝜃 = 2π rad

Example 8.3

𝑣 = 24.1 m/s

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𝑣 = 𝑟𝑒𝑟 + 𝑟𝜃 𝑒𝜃 + 𝑧𝑘 𝑣 = 0 + 2.4𝑒𝜃 − 1.2𝑘

𝑣 = 2.68 m/s

𝑎 = 𝑟 − 𝑟𝜃 2 𝑒𝑟 + 𝑟𝜃 + 2𝑟𝜃 𝑒𝜃 + 𝑧𝑘 𝑎 = −14.4𝑒𝑟 − 4.0𝑒𝜃 + 2.0𝑘 𝑎 = 15.1 m/s2

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2017/03/29

Example 8.4

MEC2023F n

During a portion of a vertical loop, an airplane flies in an arc of radius 𝜌 = 600 m with a constant speed 𝑣 = 400 km/h. When the airplane is at A, the angle made by v with the horizontal is 𝛽 = 30°, and radar tracking gives 𝑟 = 800 m and 𝜃 = 30°. Calculate 𝑣𝑟 , 𝑣𝜃 , 𝑎𝑟 and 𝜃 for this instant.

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θ

𝑎

Homework

MEC2023F

r t

Engineering Mechanics: Dynamics Meriam & Kraige (7ed)

2/167 2/169 2/171

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