CURVILINEAR MOTION: CYLINDRICAL...

Dynamics, Fourteenth EditionR.C. Hibbeler

Copyright ©2016 by Pearson Education, Inc.All rights reserved.

Today’s Objectives:

Students will be able to:

1. Determine velocity and

acceleration components

using cylindrical

coordinates.

CURVILINEAR MOTION: CYLINDRICAL COMPONENTS



1. In a polar coordinate system, the velocity vector can be

written as v = vrur + vθuθ = rur + rquq. The term q is called

A) transverse velocity. B) radial velocity.

C) angular velocity. D) angular acceleration.

. . .

2. The speed of a particle in a cylindrical coordinate system is

A) r B) rq

C) (rq)2 + (r)2 D) (rq)2 + (r)2 + (z)2

. .

. . . . .

READING QUIZ



A cylindrical coordinate

system is used in cases

where the particle moves

along a 3-D curve.

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?

APPLICATIONS



The cylindrical coordinate

system can be used to describe

the motion of the girl on the

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?

APPLICATIONS (continued)



We can express the location of P in polar coordinates as r = r ur.

Note that the radial direction, r, extends outward from the fixed

origin, O, and the transverse coordinate, q, is measured counter-

clockwise (CCW) from the horizontal.

CYLINDRICAL COMPONENTS(Section 12.8)



..

. .

Thus, the velocity vector has two components: r,

called the radial component, and rq called the

transverse component. The speed of the particle at

any given instant is the sum of the squares of both

components or

v = (r q )2 + ( r )2

The instantaneous velocity is defined as:

v = dr/dt = d(rur)/dt

v = rur + rdur

dt.

Using the chain rule:

dur/dt = (dur/dq)(dq/dt)

We can prove that dur/dq = uθ so dur/dt = quθ

Therefore: v = rur + rquθ

.

. .

VELOCITY in POLAR COORDINATES)



After manipulation, the acceleration can be

expressed as

a = (r – rq 2)ur + (rq + 2rq )uθ

.. . .. . .

.

The magnitude of acceleration is a = (r – rq 2)2 + (rq + 2rq ) 2

The term (r – rq 2) is the radial acceleration

or ar .

The term (rq + 2rq ) is the transverse

acceleration or aq .

..

.. . .

. . ... ..

The instantaneous acceleration is defined as:

a = dv/dt = (d/dt)(rur + rquθ). .

ACCELERATION (POLAR COORDINATES)



If the particle P moves along a space

curve, its position can be written as

rP = rur + zuz

Taking time derivatives and using

the chain rule:

Velocity: vP = rur + rquθ + zuz

Acceleration: aP = (r – rq 2)ur + (rq + 2rq )uθ + zuz

.. . .. . .

. . .

..

CYLINDRICAL COORDINATES



EXAMPLE: Cylindrical Robot Arm



Use a polar coordinate system and related

kinematic equations.

Given: The platform is rotating such

that, at any instant, its angular

position is q = (4t3/2) rad, where

t is in seconds.

A ball rolls outward so that its

position is r = (0.1t3) m.

Find: The magnitude of velocity and acceleration of the

ball when t = 1.5 s.

Plan:

EXAMPLE



Solution:

𝑟 = 0.1𝑡3, ሶr = 0.3 t2, ሷr = 0.6 t

𝜃 = 4 t3/2, ሶ𝜃 = 6 t1/2, ሷ𝜃 = 3 t−1/2

At t=1.5 s,

r = 0.3375 m, ሶr = 0.675 m/s, ሷr = 0.9 m/s2

𝜃 = 7.348 rad, ሶ𝜃 = 7.348 rad/s, ሷ𝜃 = 2.449 rad/s2

Substitute into the equation for velocity

v = r ur + rq uθ = 0.675 ur + 0.3375 (7.348) uθ

= 0.675 ur + 2.480 uθ

v = (0.675)2 + (2.480)2 = 2.57 m/s

. .

EXAMPLE (continued)



Substitute in the equation for acceleration:

a = (r – rq 2)ur + (rq + 2rq)uθ

a = [0.9 – 0.3375(7.348)2] ur

+ [0.3375(2.449) + 2(0.675)(7.348)] uθ

a = – 17.33 ur + 10.75 uθ m/s2

a = (– 17.33)2 + (10.75)2 = 20.4 m/s2

.. . .. . .

EXAMPLE (continued)



1. If r is zero for a particle, the particle is

A) not moving. B) moving in a circular path.

C) moving on a straight line. D) moving with constant velocity.

.

2. If a particle moves in a circular path with constant velocity, its

radial acceleration is

A) zero. B) r .

C) − rq 2. D) 2rq .

..

. . .

CONCEPT QUIZ



Use cylindrical coordinates.

Given: The arm of the robot is

extending at a constant rate

ሶ𝑟 = 1.5 ft/s when r = 3 ft,

z = (4t2) ft, and q = (0.5 t) rad,

where t is in seconds.

Find: The velocity and acceleration

of the grip A when t = 3 s.

Plan:

GROUP PROBLEM SOLVING



Solution:

When t = 3 s, r = 3 ft and the arm is extending at a constant

rate ሶ𝑟 = 1.5 ft/s. Thus ሷ𝑟 = 0 ft/s2

𝜃 = 1.5 t = 4.5 rad, ሶ𝜃 = 1.5 rad/s, ሷ𝜃 = 0 rad/s2

z = 4 t2 = 36 ft, ሶz = 8 t = 24 ft/s, ሷz = 8 ft/s2

Substitute in the equation for velocity

v = r ur + rq uθ + z ur

= 1.5 ur + 3 (1.5) uθ + 24 uz

= 1.5 ur + 4.5 uθ + 24 uz

Magnitude v = (1.5)2 + (4.5)2 + (24)2 = 24.5 ft/s

. . .

GROUP PROBLEM SOLVING (continued)



Acceleration equation in cylindrical coordinates

.. . .. . . ..a = (r – rq 2)ur + (rq + 2rq)uθ + zuz

= {0 – 3 (1.5)2}ur +{3 (0) + 2 (1.5) 1.5 } uθ + 8 uz

a = [6.75 ur + 4.5 uθ + 8 uz] ft/s2

a = (6.75)2 + (4.5)2 + (8)2 = 11.4 ft/s2

GROUP PROBLEM SOLVING (continued)



2. The radial component of acceleration of a particle moving in

a circular path is always

A) negative.

B) directed toward the center of the path.

C) perpendicular to the transverse component of acceleration.

D) All of the above.

1. The radial component of velocity of a particle moving in a

circular path is always

A) zero.

B) constant.

C) greater than its transverse component.

D) less than its transverse component.

ATTENTION QUIZ

CURVILINEAR MOTION: CYLINDRICAL...

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