Dynamics 14th edition by r c hibbeler section 12 8

In-Class Activities: • Velocity Components • Acceleration Components CURVILINEAR MOTION: CYLINDRICAL COMPONENTS... A cylindrical coordinate system is used in cases where the particle mov

Today’s Objectives:

Students will be able to:

using cylindrical coordinates

In-Class Activities:

• Velocity Components

• Acceleration Components

CURVILINEAR MOTION: CYLINDRICAL COMPONENTS

1 In a polar coordinate system, the velocity vector can be written as v = vrur + vθuθ = rur + rθuθ The term

θ 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

C) (rθ)2 + (r)2 D) (rθ)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, θ, is measured counter-clockwise (CCW) from the horizontal

CYLINDRICAL COMPONENTS

(Section 12.8)

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 θ )2 + ( r )2

The instantaneous velocity is defined as:

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

v = rur+ r dur

dt

.

Using the chain rule:

dur/dt = (dur/dθ)(dθ/dt)

Therefore: v = rur + rθuθ

VELOCITY in POLAR COORDINATES)

After manipulation, the acceleration can be expressed as

a = (r – rθ 2)ur + (rθ + 2rθ )uθ

.

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

The term (r – rθ2) is the radial acceleration or ar

The term (rθ + 2rθ ) is the transverse acceleration or aθ

The instantaneous acceleration is defined as:

a = dv/dt = (d/dt)(rur + rθuθ) .

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 + rθuθ + zuz

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

CYLINDRICAL COORDINATES

Use a polar coordinate system and related kinematic equations.

Given: The platform is rotating such that, at any instant, its

angular position is θ = (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

, ,

4 t3/2, 6 , 3

At t=1.5 s,

r 0.3375 m, 0.675 m/s, 0.9 m/s2

7.348 rad, 7.348 rad/s, 2.449 rad/s2

Substitute into the equation for velocity

v = r ur + rθ 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 – rθ 2)ur + (rθ + 2rθ)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

.

CONCEPT QUIZ

Use cylindrical coordinates.

Given: The arm of the robot is extending at a constant rate

= 1.5 ft/s when r = 3 ft,

z = (4t2) ft, and θ = (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 = 1.5 ft/s Thus ft/s2

1.5 t 4.5 rad, 1.5 rad/s, 0 rad/s2

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

Substitute in the equation for velocity

v = r ur + rθ 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 – rθ 2)ur + (rθ + 2rθ)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

End of the Lecture Let Learning Continue