Curvilinear Motion and Polar Coordinates
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Transcript of Curvilinear Motion and Polar Coordinates
KINEMATICS OF PARTICLES
Kinematics of particles is the study about the motion of particles - specifically todetermine their d i s p I a c e m e nt, v e I o c ity and a c c e I e r at i o n.
o Displacement: position relative to a reference point or coordinate.
o velocity: rate of change of displacement with respect to time.
o Acceleration: rate of change of velocity with respect to time.
Motion :
The motion of a parlicleand c urv il ine ar mot i o n.
o Rectilinear motion:
c Curvilinear motion:
may take place in two main forms, namely rectilinear motion
The particle moves in a straight line.
The particle moves at random - i.e. a combination of linearand rotation.
Rectilinear Motion
In this motion, displacement, velocity anddepending on the direction of motion
Motion inX-direction :
Motion in l-direction :
Displacement :
Velocity:
Acceleration :
acceleration are indicated in terms x or !,
Displacement:
Velocity:
Acceleration :
x
v*= iar= t
vv, = j'
an= !
Curvilinear Motion
In this motion, displacement, velocity and acceleration are measured by means ofCartesian coordinates (x-y) or Polar coordinates (r-0).
Cartesian coordinates
The displacement of the particle isresolved into two components, i.e. onealong the OX axis and the other alongthe OIaxis, as shown on the right.
The velocity of the particle, vo, isresolved into two components, i.e. onealong the OX axis and the other alongthe OY axis, as shown below.
lr= x
vv = j'
Likewise for the acceleratio,n of the particle :
o*= i
A-.: iyJ
al
"o='1 *utr
o?: o1 *
Polar coordinates
The displacement of the particle isexpressed in terms of the radius r and
the angle 0 measured from the OX axis,as shown on the right.
The velocity of the particle is expressed
in terms of a rqdial comPonent and atransverse component, as shown below.
lr=f
ve=r0
)))vi=vi +v-e
Similarly, the acceleration of the particle is expressed
and a transverse component, as shown below.in terms of a radial component
ar= i - '62
oe= r0 + 2i0
azp: a? + a3
Relstionship between Cartesian and porar coordinates
o Displacement:
It can be seen from the figure given on theright that :
x-Y=
r.cos0
r. sin 0
Also:
And :
f=
e- :Vtan 'z-
x
o Velocity:
The velocity of particle p, ( v o ), may be resolved into :
o eitherthe Cartesiancomponents: v* and v, , or
o the Polar components : v, and v,
as shown in the figure below :
*'+y'
From the above figure, it can be seen that :
o r, : vr.cosO + vr.sinO = *.cos0 + y.sinO (1)
o ve = vr.cos0 - vr.sin0 = y.cos0 - *.sinO (2)
Now, from displacement :
x = r.cosO and .lz = r.sin0
* = r.coso - r.sin0.6 (i)
& -tr = r.sin0 + r.cos0.6 Gi)
Therefore, substituting (i) and (ii) into equations (l) and (2) above will produce :
o v, = ( r.cos0 - r.sin0.6 ).cos0 + ( r.sinO + r.cos0.O ).sinO
: r.cos20-rO.sinO.cosO + r.sin2g+rO.cos0.sin0 = i (l):
o v0 = ( r.sin0 + r.cos0.O ).cosO ( r.cosO - r.sin0.0 ).sin0
= r.sinO.cosO + r6.cos2e r.cosO.sinO + rO.sin20 = rd e)
o Acceleration:
The acceleration of particle P, (ap), can be resolved into :
" eitherthe Cartesiancomponents: o, and a, , or
o the Polar components : a, antd au
The diagram for these acceleration components will be identical to the diagram for thevelocities above. Therefore, it will be seen that :
o a, = ar.cosO + ar.sinO = i.cosO + /.sin0 (3)
o ao = ar.cosO - ar.sinO = /.cos0 - x.sinO (4)
Now differentiating (i) and (ii) above will give :
I = r.cosO - r.sin0.6 - r.sin0.6 - r.cos0.02 - r.sin0.6 (iii)
& i = r.sin0+r.cos0.6 + r.cos0.O-r.sin0.02 +r.cos0.6 (iv)
Substituting (iii) and (iv) into equations (3) and (a) for accelerations, and simpliSing,will produce :
o o,=i;-162 (3)
o ao=16+2i6 g)
Summary
The polar components for the velocity and acceleration in curvilinear motion may besummarised as shown below :
o
o
lr=r
vo=,6o ar=i-162
as= 16 +2i6o
Special Cases
o Linear Motion:
When the motion is linear, angle 0 remains constant, giving 0 = 0 and 6 = 0. Hencethe equations above will reduce to :
o Circular Motion:
When the motion is circular, radius r remains constant, giving r = 0 and i = 0. Hencethe equations above will reduce to :
o
o
vr=0
vo = ,6o
o
o, = -'6'ae = rd
or - ra2
or ra
ovr=r
o Vg=0
ar=f
ae=0
o
o