ONLINE: MATHEMATICS EXTENSION 2
Topic 6
6.61
MECHANICS
MOTION IN A CIRCLE
COMPLEX NUMBERS
We can use complex numbers to mathematically investigate the motion of a particle in a
circle.
Let the X axis be the real axis and the Y axis be the imaginary axis. The position of the
particle is given by the complex number z as a vector
z  x  i y  R  cos  i sin    R ei 
where the angular displacement  is a function of time     t  . The magnitude of z is
z R
and the angle of the complex vector z makes with the real axis is the argument of z
Arg  z   
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For motion in a circle the radius is constant
dR d 2 R
 2 0
dt
dt
RR0
The velocity v of the particle is found by differentiating the position of the particle z
with respect to time t.
dz
z
dt
d
 d  i 
i
v   R ei    R  i
 e  i R e
dt
dt


vi z
v
where the angular speed is  
d
 .
dt
The magnitude of the velocity is v  R  . Multiplication by i produces a rotation of a
complex vector by  /2 rad. Therefore z and i z are at right angles to each other. Hence,
the displacement vector z and velocity vector v are always perpendicular to each other.
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The acceleration is found by differentiating the velocity with respect to time
dv
v
dt
d
d
d i  
 d i 
a   i  z    i  R ei    i R 
e  i
e 
dt
dt
dt
 dt

d
ai
R ei    2 R ei 
dt
d  d 2
 d 
2
i /2
2
a  e i  / 2  
z


z

e

z


z


 2 
 

dt
dt
 dt 
a
where the angular acceleration is  
The first term
e
i /2
d
.
dt
  ddt  z corresponds to the change in speed of the particle and the
direction is perpendicular to the displacement vector. This term is the tangential
acceleration aT. The magnitude of the tangential acceleration is
d  dv
 d 
aT   e i  / 2  

z  R
dt
dt
 dt 
The second term  2 z corresponds to the change in direction and it is in a direction
opposite to the displacement vector. This is the radial acceleration aR. The magnitude of
the radial acceleration is
aR   2 z   2 R 
v2
R
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