CURVILINEAR MOTION: CYLINDRICAL COMPONENTS

CURVILINEAR MOTION: CYLINDRICAL COMPONENTS

Today’s Objectives: In-Class Activities:
Students will be able to: • Check Homework
1. Determine velocity and • Reading Quiz

acceleration components
using cylindrical
coordinates.

• Applications

• Velocity Components

• Acceleration Components

• Concept Quiz

• Group Problem Solving

• Attention Quiz

READING QUIZ

1. In a polar coordinate system, the ve.locity vector ca. n be
written as v = vrur + vθuθ = r.ur + called
rquq. The term q is

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) . + (r.)2 D) . + (r. )2 + (z.)2

(rq)2 (rq)2

APPLICATIONS

The 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
know if the package will fly
off the ramp?

CYLINDRICAL COMPONENTS
(Section 12.8)

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.

VELOCITY in POLAR COORDINATES)

The instantaneous velocity is defined as:

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

v = .rur + r dur

dt

Using the chain rule:

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

We can prove th.at dur/.dq = uθ so dur/dt = quθ

Therefore: v = rur + rquθ .

Thus, the velocity vector has two c.omponents: 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 . )2 + ( .r )2

q

ACCELERATION (POLAR COORDINATES)

The instantaneous acceleration is defined as:

a = dv/dt = (d/dt)(.rur + .

rquθ)

After manipulation, the acceleration can be
expressed as

The teram=(.r(..r–. –rqr.q.2)2)ius rth+e(rr.a.qd+ia2l .raqc.)cueθleration

or ar .

The term .. + 2r.q.) is the transverse

(rq

acceleration or aq .

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

rq (rq

CYLINDRICAL 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 = r.ur + . + z. uz
rquθ

Acceleration: aP = (.r. – . + .. + 2.rq. )uθ + .z.uz
rq2)ur (rq

EXAMPLE

Given: A car travels. along acircular path.

r..= 300 ft, q = 0.4 (rad/s),
q = 0.2 (rad/s2)

Find: Velocity and acceleration
Plan: Use the polar coordinate system.

Solution: . .r. . ..
r
r = 300 ft, = = 0, and q = 0.4 (rad/s), q= 0.2 (rad/s2)

Substitute in the equati.on for velocity
v = r.
ur + rq uθ = 0 ur + 300 (0.4) uθ

v = (0)2 + (120)2 = 120 ft/s

EXAMPLE
(continued)

Substitute in the equation for acceleration:

a = (.r. – . 2)ur + .. 2.rq.)uθ

rq (rq +

a = [0 – 300(0.4)2] ur + [300(0.2) + 2(0)(0.4)] uθ

a = – 48 ur + 60 uθ ft/s2

a = (– 48)2 + (60)2 = 76.8 ft/s2

CONCEPT QUIZ

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 .

. D) 2r.q. .

C) − rq 2.

GROUP PROBLEM SOLVING

Given: The car’s speed is constant at
1.5 m/s.

Find: The car’s acceleration (as a
vector).

Plan:

Hint: The tangent to the ramp at any point is at an angle

f = tan-1( 12 ) = 10.81°
2p(10)

Also, what is the relationship between f and q?

Plan: Use cylindrical coordinates. Since r is constant, all
derivatives of r will be zero.

GROUP PROBLEM SOLVING (continued)

Solution: Since r i.s constant, the velocity o. nly has 2 components:
z=
vq = rq = v cos f and vz = v sin f

Therefore: . = ( v cosf ) = 0.147 rad/s

q r

..
q=0

zv..z = z. = v sinf = 0.281 m/s
= 0
. ..
r=r=0

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

rq (rq

.
a = (-rq 2)ur = -10(0.147)2ur = -0.217ur m/s2

ATTENTION QUIZ
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.

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.