kinematics of a particles general curvilinear motion, Exercises of Mechanics

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Chapter 1(cont.)

Kinematics of Particles

Curvilinear Motion

►Cartesian Coordinates

►Intrinsic Coordinates

►Polar Coordinates

II

Prof. Imam Morgan ( 2 )

5 Curvilinear Motion

Particle moving along a curve other than a straight line is in

curvilinear motion. Indeed, in the shown photo, each plane

performs aspace curvilinear motion. We are interest to

specify, completely, this motion and to determine the

kinematic quantities that define the motion

Prof. Imam Morgan ( 3 )

a



r



v



P

Path

O

General

Objectives r Positionvector  r r   t 

    ... 

  

tangenttothe path

v Velocityvector v v t

    ... 

  

nottangenttothe path

a Accelerationvector a a t

    ... 

► How can we express each

vector at any time.

► How can derive the relation(s)

between them.

► The physical meaning of each

vector (why? and how?)

Kinematic

Parameters

Repeated Slide

!!

Prof. Imam Morgan ( 4 )

Curvilinear Motion

(General Discussion)

Curvilinear motion occurs when the particle moves along a curved path.

Since this path is often described in three dimensions, vector analysis

(mainly components) will be used to formulate the particle’s position,

velocity, and acceleration.

5.1 Position Vector, Velocity Vector, and Acceleration vector

A (^) Position Vector

► Position vector of a particle at

time t is defined by a vector

between origin O of a fixed

reference frame and the

position occupied by particle.

Prof. Imam Morgan ( 5 )

Consider particle which occupies

positionP defined by at timet

and P’ defined by at t + Dt,

r



r 



s curvilineardisplacement

r displacement vector

...

...

D

D



Theinstantaneous velocity of the particle at

timet is obtained by considering thatP’

approachesP.

So, becomes shorter and shorter,

Hence, in the limit att = 0 it tends to be in

tangent direction.

r

 D

B (^) Velocity Vector

Prof. Imam Morgan ( 6 )





D

D







D

D



D 

D

dt

ds

t

s v

dt

dr

t

r v

t

t

0

0

lim

lim

 



instantaneous velocity ( vector )

instantaneous speed ( scalar )

is tangent to

the path

v



Prof. Imam Morgan ( 7 )

C (^) Acceleration Vector

Consider velocity of particle at

timet and velocity at t + Dt,

v



v





v



v



D v^  





D

D



D  dt

dv

t

v a t

 



0

lim

instantaneous acceleration ( vector )

tothepath

a isnot tangent



Prof. Imam Morgan ( 8 )

Finally

doesnotgive thedirection of motion

not tangent tothepath(inconcaveside)

...acceleration vector

gives thedirectionof motion

tangent tothepath

...velocity vector

... position vector

dt

dv a

a

dt

dr v

v

r



















Prof. Imam Morgan ( 9 )

►In this case, the position vector of particleP is

given by its rectangular components,

r xi yj zk

   

  

► Velocity vector ,

v i v j v k

k xi yj zk dt

dz j dt

dy i dt

dx v

x y z

 ^ 













   

  

     

The total velocity v is tangent to the path.



5.2 Rectangular Components

Now, we shall express each vector in terms of its components for different cases;

namely: Rectangular Components

Tangential and Normal components

Radial and Transversal components

5.

5.

5.

Prof. Imam Morgan ( 10 )

► Acceleration vector ,

a i a j a k

k xi yj zk dt

d z j dt

d y i dt

d x a

x y z

 ^ 













   

  

     

2

2

2

2

2

2

a

 in concave side

Prof. Imam Morgan ( 11 )

Projectile

Path

Vertical

Plane

Reference

Line

P

Firing (initial)

Velocity

Reference Plane

Projectile

Motion ►An important application

of Cartesian coordinates is

that for study of motion

of a particle that moves in

a constant acceleration

field: (Projectile Motion)

►Basic Assumptions:

● g = constant (near

surface of the earth)

● Neglect air resistance

● The projectile is

considered as a

particle.

Important application

g = constant

Prof. Imam Morgan ( 12 )

vo

vx=?

vy=?

y(t)=?

x(t)=?

α

Basic Relations

Given : vo ≡ initial velocity α ≡ inclination of vo

w.r.to horizontal

Required : Position ……… x (t) , y (t) Velocity ……… vx (t) ,vy (t)

Equation of the path y = y (x)

 

    2

1

0

x v cos t

v v cos

x

o

x o

..









  

    4 2

1

3

2 y v sin t gt

v v sin gt

y g

o

y o

..

 

 







Eliminating t from (2) and (4) we get the path

x

y

O

 

  5 2 cos

tan

2 2 2

x v

g y x

o^ 

  

Parabolic Path

Prof. Imam Morgan ( 13 )

Important Note

►The equations (1) through (5) can be used to solve any projectile’s

motion if the following conditions are satisfied:

● The originO is taken at the firing point.

● x-axis is horizontal (in direction of firing).

● y-axis is vertically upward.

x

y

α = 0

x

y

α = - 15 o

4 km

1 km^ x

y

x

y

Prof. Imam Morgan ( 14 )

For the shown projectile, find in terms ofvo and α the following:

a- The time of flight fromA toC.

b- The rangeR.

c- The maximum heightHmax

C

B

O ≡ A

Hmax

R

x

y

vo

α

Example (5)

Prof. Imam Morgan ( 15 )

a- The time of flight fromA toC.

this time is denoted asT = t A→C.

 

 

g

v sin T

v sin T gT

useequation ; and substitutey and t T

at C: y

o

o





2

2

1 0

4 0

0

2



 

 



b- The rangeR.

you can use equation (2) by substitutingx = R andt = T.

 

 

 

g

v sin R

g

v sin v cos

R v cos T

o

o o

o



 



2

2

2 





C

B

A

vo

α

R

Hmax

y

x

Prof. Imam Morgan ( 16 )

vo

α = 45 o

Rmax

α > 45 o

vo

α < 45 o

vo

The maximum rang

Rmax is obtained when

α = 45 o^ in condition

that vo is kept constant

g

v R

o max

2 

 

g

v R

o sin^2 

2



Prof. Imam Morgan ( 17 )

c- The maximum heightHmax.

at B we have the condition thatvy = 0

So, use equation (3) to get the time of flight fromA toB.

 

g

v H

g

v

g

v H v

substitute in

g

v t

v gt

o

o o o

o A B

o A B

sin

sin

sin 1 sin

4 ;

sin

0 sin

2 2

max

2

max



  































 



 





…… (6)

C

B

A

vo

α

R

Hmax

y

x

The equations forT,R, and

Hmax are valid only for this

Figure.

Important

Prof. Imam Morgan ( 18 )

C

B

O ≡ A

x

y

y

E

D

vE

vD

Important Note

►The projectile will have the same velocity, as it passes through the same

height y: vD= vE. Moreover, the velocity aty can be obtained from:

v vo 2 g y

2 2   Proof is recommended

vo

α

Prof. Imam Morgan ( 19 )

A projectile is fired from the edge of a 150 m cliff with an initial velocity

vo = 180 m/s as shown.

a) Determine the horizontal distance (R) fromA toC.

b) Calculate the time of flight fromA toC.

c) Find the maximum height (hmax) above the lower horizontal plane.

R

hmax

C

B A

R

hmax

C

B A

y

x

vo= 180 m/s α = 30 o

Example (6)

Prof. Imam Morgan ( 20 )

a) Determine the horizontal distance (R) fromA toC.

 

   

 

 

R. m

R. m or, R. m refused

. R. R

R cos

. Rtan

x R and y

Useequationof the path ,whereatC:

30977

30977 239 8

2019 10 0577 150 0

2180 30

981 150 30

150

5

4 2

2 2 2

 

 

   

  

 



◄

b) Calculate the time of flight fromA toC.

 

   

 

T. sec

T. sec or, T. sec refused

T. T.

sin T. T

Inequation substitute y

1991

1991 154

1837 306 0

981 2

1 150 180 30

4 150

2

2

 

 

  

  



◄

R

hmax

C

B A

y

x

vo= 180 m/s α = 30 o

One can use eq. (2)

withx = 3097.7 m

Prof. Imam Morgan ( 21 )

c) Find the maximum height (hmax) above the lower horizontal plane.

   

 

h m

.

sin h

TheheightH above x axisis obtained before:

at B: v

max

max

max

y

563

150 413

2 981

180 30 150

0

2 2

 

 

  





◄

R

hmax

C

B A

y

x

vo= 180 m/s α = 30 o

Prof. Imam Morgan ( 22 )

Example (7)

A projectile is fired with an initial

velocity of 800 ft/s at a targetB

located 2000 ft above the gunA at

horizontal distance of 12000 ft.

Neglecting air resistance, determine

the value of the firing angle α.

Solution:

We can use directly equation of

the path:

2 2 2 2 cos

tan x v

g y x

o^ 

  

Where, at B:

x = 12000 ft, y = 2000 ft

Prof. Imam Morgan ( 23 )

 

 

2 (^2 2)

2 800 cos

2000 12000 tan



  

Knowing that:

 



2 2 2

sec 1 tan cos

  

then, 2000 = 12000 tan α – 3622. 5 [ 1 + tan^2 α ]

3622.5 tan^2 α – 12000 tan α + 5622. 5 = 0

tan α 1 = 0. 565 or, tan α 2 = 2. 748

α 1 = 29. 5 o^ or, α 2 = 70 o

So, the target will be hit if either of these two firing

angles is used (two possible solutions)

Recommended :

Calculate the time of flight

fromA toB in each angle.

Prof. Imam Morgan ( 24 )

This type of coordinates is used

wheneverthe path of the particle

is known (well defined).

5.3 Tangential and Normal Components (Intrinsic or,

Natural Coordinates)

or

Prof. Imam Morgan ( 25 )

s = s(t)

Position

Po O

P

s(t)

Plane Path

Theposition of the particle on the shown curved plane path is

defined by the curvilinear distance s covered by the particle

during time t.

The reference point O is considered as a point on the path.