Objective• To investigate particle motion along a curved path “Curvilinear

Motion” using three coordinate systems

– Rectangular Components• Position vector r = x i + y j + z k

• Velocity v = vx i + vy j + vz k (tangent to path)

• Acceleration a = ax i + ay j +az k (tangent to hodograph)

– Normal and Tangential Components• Position (particle itself)

• Velocity v = ut (tangent to path)

• Acceleration (normal & tangent)

– Polar & Cylindrical Components

nt uua2

22

2/32

/

)/(1

dxyd

dxdy

Curvilinear Motion: Cylindrical Components

• Section 12.8

• Observed and/or guided from origin or from the center

• Cylindrical component

• Polar component “plane motion”

zandr ,,andr

Application: Circular motion but observed and/or controlled from the center

Polar Coordinates

• Radial coordinate r

• Transverse coordinate

• and r are perpendicular

• Theta in radians

• 1 rad = 180o/• Direction ur and u

Position

• Position vector

• r = r ur

Velocity• Instantaneous velocity

= time derivative of r

• Where

rr rr uu rv

uu r

uuv rr

randrr

uuv rr r

ru r r

velocitye transversv

velocityradial vr

Velocity (con.)

• Magnitude of velocity

• Angular velocity • Tangent to the path• Angle =

22 )()( rr

)(tan 1

r

Acceleration• Instantaneous acceleration = time derivative of v

uuv rr r

2 rrar

uuuuu va rrrrr rr

ruu

uua aa rr

rra 2

uu r

u)2(u)(a 2 rrrr r

Acceleration (con.)

• Angular acceleration

• Magnitude

• Direction “Not tangent”• Angle

)/(// 22 dtddtddtd

222 )2()( rrrra

)(tan 1

ra

a

Cylindrical Coordinates

• For spiral motion cylindrical coordinates is used r, , and z.

• Position

• Velocity

• Acceleration

zrp zr uur

zr zrr uuuv

zr zrrrr uu)2(u)(a 2

Time Derivative to evaluate

• If r = r(t) and t)

• If r = f( use chain rule

andrr ,,,

)68(4 32 ttr 2248 ttr

tr 488

25r 10r

])()([10 r

1010 2 r

• The slotted fork is rotating about O at a constant rate of 3 rad/s. Determine the radial and transverse components of velocity and acceleration of the pin A at the instant = 360o. The path is defined by the spiral groove r = (5+ in., where is in radians.

2

2

in/s 0 in/s 3

in 75

rrr

2rad/s 0rad/s 3rad 2360 o

in/s955.03

rvr

in/s21)3(7 rv

222 in/s63)3(70 rrar

2in/s73.5)3)(3

(202

rra

Problem

Example 12-20

and find

180at ft/s30

ft/s4

ft)cos1(5.0

o2

a

v

r

)cos1(5.0 r

)sin(5.0r )sin(5.0)()(cos5.0 r

2

o

50 0 ft 1

180at

θ.-rrr

rad/s4 4)1()0()()( 2222 rr

30)]4)(0(21[])4(1)4(5.0[)2()( 2222222 rrrra

2rad/s18

0/84*22

*2 rsmrmr

)56.12()8( 22

A collar slides along the smooth vertical spiral rod, r = (2) m, where is in radians. If its angular rate of rotation is constant and equal 4 rad/s, at the instant = 90o. Determine

-The collar radial and transverse component of velocity -The collar radial and transverse component of acceleration.

-The magnitude of velocity and acceleration

Problem

222 rrr

2/0/42

sradsradrad

m/s8rvr

m/s56.12)4( rv

222 m/s24.50)4(0 rrar2m/s64)4)(8(202

rra

)64()24.50( 22 a