The description of the Plane Curvilinear Motion by the normal-tangential (n-t) coordinates & the

Polar Coordinates

Lecture IV

t

nPA

PB

PC

t

n

t

nPath

y

x

Pr

r

Path

O

Polar coordinate

s

Normal-Tangential coordinates

Plane Curvilinear Motion – Normal-Tangential (n-t)

Coordinates

t

nPA

PB

PC

t

n

t

nPath

Here, the curvilinear motions measurements are made along the tangent (t) and the normal (n) to the path.

n-t coordinates are considered to move along the path with the particle.

The positive direction of the normal (n) always points to the center of curvature of the path; while the positive direction of the tangent (t) is taken in the direction of particle advance (for convenience).

et & en are the unit vectors in t-direction and n-direction, respectively.

(n-t) Coordinates - VelocityNote: is the radius of curvature and d is the increment in the angle (in radians)

(after dt)

tt evv

dt

d

dt

dsvt

Note: as mentioned before that the velocity vector v is always tangent to the path; thus, the velocity has only one component in the n-t coordinates, which is in the t-direction. This means that vn = 0.

Its magnitude is:

(n-t) Coordinates - Acceleration

Note: et, in this case, has a non-zero derivative, since it changes its direction. Its magnitude remains constant at 1.

tttt

ttevev

dt

evddt

d

a

va

?

t

ntn

t

n

t

nt

vdtds

eeedt

d

dt

ed

dt

ed

ededed

:But

:by Dividing

)after dt(

Note: the vector det , in the limit, has a magnitude equal to the length of the arc |et|d=d. The direction of det is given by en.

Thus,

nt

tt ev

ev

2

a

(n-t) Coordinates – Acceleration (Cont.)

22

22

tn

tn

tt

aaa

vv

a

sva

Its magnitude is:

Notes:

-an always directed toward the center of curvature.

-at positive if the speed v is increasing and negative if v is decreasing.

= , thus an = 0

(n-t) Coordinates – Circular Motion

tt

n

tt

t

vrr

va

rva

rv

22

For a circular path: = r

n-t Coordinates Exercises

Exercise # 1

2/97: A particle moves in a circular path of 0.4 m radius. Calculate the magnitude a of the acceleration of the particle (a) if its speed is constant at 0.6 m/s and (b) if its speed is 0.6 m/s but is increasing at the rate of 1.2 m/s each second. .

Exercise # 22/101: The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.

Exercise # 32/110: Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates

for the instant when = 60° if = 2.00 rad/s

and .. = 2.45 rad/s2.

Exercise # 42/118: The design of a camshaft-drive system of a four cylinder automobile engine is shown. As the engine is revved up, the belt speed v changes uniformly from 3 m/s to 6 m/s over a two-second interval. Calculate the magnitudes of the accelerations of points P1 and P2 halfway through this time interval.

Exercise # 52/128: The pin P is constrained to move in the slotted guides which move at right angles to one another. At the instant represented, A has a velocity to the right of 0.2 m/s which is decreasing at the rate of 0.75 m/s each second. At the same time, B is moving down with a velocity of 0.15 m/s which is decreasing at the rate of 0.5 m/s each second. For this instant determine the radius of curvature of the path followed by P.

Plane Curvilinear Motion – Polar Coordinates

Here, the curvilinear motions measurements are made by the radial distance (r) from a fixed pole and by an angular measurement () to the radial line.

The x-axis is used as a reference line for the measurement of . er & e are the unit vectors in r-direction and -direction,

respectively.

y

x

Pr

r

Path

O

Polar Coordinates – Position & Velocity

tt evv

Note: from (b), der is in the positive -direction and de in the negative r-direction

rerr

The position vector of the particle:

rr

terer

dt

erd

dt

d v r

v

The velocity is:?

)after dt(

)after dt(

rr

r

r

rr

r

r

ee edt

dθ

dt

ed

eeedt

dθ

dt

ed

dt

ed

ededed

ed

ededed

:by Dividing

Polar Coordinates – Velocity (Cont.)

erer r v

Thus, the velocity is:

22

vvv

rv

rv

r

r

Its magnitude is:

Due to rotation of r

Due to the rate at which the vector stretches

Polar Coordinates - Acceleration

ererererer

dt

ererd

dt

d

rr

r

a

va

Rearranging,

errerr r 2a 2

22

2

2

aaa

rra

rra

r

r

Its magnitude is:

Coriolis acceleration

Centripetal acceleration

Polar Coordinates – Circular Motion

ra

ra

rv

v

r

r

2

0For a circular path: r = constant

Note: The positive r-direction is in the negative n-direction, i.e. ar = - an

Polar Coordinates

Exercises

Exercise # 6 2/131: The boom OAB pivots about point O, while

section AB simultaneously extends from within section OA. Determine the velocity and acceleration of the center B of the pulley for the following conditions: = 20°,

= 5 deg/s, = 2 deg/s2, l = 2

m, l = 0.5 m/s, l

= -1.2 m/s2. The quantities l

and l

are the first and second time derivatives, respectively, of the length l of section AB.

Exercise # 72/133: The position of the slider P in the rotating slotted arm OA is controlled by a power screw as shown. At the instant represented, . = 8 rad/s and .. = -20 rad/s2. Also at this same instant, r = 200 mm, r. = -300 mm/s, and r.. = 0. For this instant determine the r- and -components of the acceleration of P .

Exercise # 82/142: At the bottom of a loop in the vertical (r-) plane at an altitude of 400 m, the airplane P has a horizontal velocity of 600 km/h and no horizontal acceleration. The radius of curvature of the loop is 1200 m. For the radar tracking at O, determine the recorded values of r.. and .. for this instant.