Kinematics in 2-D

1. Position

a) Rectangular Coordinates: x(t) and y(t)

r = xx + yy ,

where x and y are constant unit vectors.

b) Polar Coordinates: r(t) and q(t)

r = rr .

Note: r  ¹  rr + qq unlike what we had in rectangular. The r dependence is obvious, but where is the q dependence? It turns out that the q dependence is in the unit vector, r. If we draw a graph, it is easy to see that we can break the unit vector, r, into rectangular components:

r  =  cos(q) x + sin(q) y .

What is the direction for the unit vector q? To make the system an orthogonal system, we need q to be perpendicular to r. We can make this happen by simply adding on 90o to the direction of r:

q  =  cos(q+90o) x + sin(q+90o) y .

We recognize that cos(q+90o) = -sin(q) and that sin(q+90o) = +cos(q) so that

q  =  -sin(q) x + cos(q) y .

Note that both unit vectors in polar coordinates, r and q, change direction as the angle q changes! This means that

r  =  r(q) and q = q(q).

If we want to show the functional dependence on time, we can write:  r(t)  =  x(t)x + y(t)y  and  r(t)  =  r(t)r(q(t)).

2. Velocity

a) Rectangular Coordinates:

v  =  dr/dt  =  d[xx + yy]/dt  =  (dx/dt)x + x(dx/dt) + (dy/dt)y + y(dy/dt)  =  vxx + vyy ,

since dx/dt = 0 and dy/dt = 0 ,  since both x and y are constants.  Note also that vx = dx(t)/dt   and  vy = dy(t)/dt .

b) Polar Coordinates:

v  =  dr/dt  =  d[rr]/dt  =  (dr/dt)r + r(dr/dt) .

Since r  =  cos(q) x + sin(q) y,   dr/dt  =  d[cos(q(t)) x + sin(q(t)) y]/dt  =  d[cos(q) x + sin(q) y]/dq * dq(t)/dt  =   -sin(q) (dq/dt) x + cos(q) (dq/dt) y .

We now define w =dq/dt, and we recall from above that   q  =  -sin(q) x + cos(q) y.  This gives the result:

dr(q(t))/dt  =  (dr/dq)(dq/dt)  =  wq  =  w(t) q(q(t))

Note in particular that (dr/dq) = q .   Using vr = dr/dt and  vq = rw, we have:

v  =  vr r + rwq   =   vr r + vq q .

These results correspond to what we found in Physics I when we looked at circular motion.

3. Acceleration

a) Rectangular Coordinates:

a  =  dv/dt  =  d[vxx + vyy]/dt  =  (dvx/dt)x + (dvy/dt)y  =  axx + ayy .

b) Polar Coordinates:

a  =  dv/dt  =  d[vrr + rwq ]/dt  =  (dvr/dt)r + vr(dr/dt) + (dr/dt)wq + r(dw/dt)q + rw(dq /dt).

In the second term we have (dr/dt) = wq .  In the third term, we have (dr/dt) = vr. In the fourth term, we have (dw/dt) = a. In the fifth term, we have

dq/dt  =  d[-sin(q) x + cos(q) y]/dt  =  -cos(q)(dq/dt)x + -sin(q)(dq/dt)y  =  -wr .

In particular we have the useful result that

dq /dq = -r.

Putting all this back together gives:

a  =  (dvr/dt)r + vrwq + vrwq + raq + rw(-w)r  =  [(dvr/dt) - w2r]r + [(2vrw)+ra]q .

(1) We recognize (dvr/dt) as the normal straight-line acceleration in the radial direction.
(2) We recognize the
w2r as the centripetal acceleration directed toward the center, or in the -r direction.
(3) We recognize r
a as the normal tangential acceleration.
(4) But what is the 2vr
w term in the q direction? We call this the Coriolis acceleration. What is this due to? Consider a merry-go-round. As you walk from near the middle out towards the end, you have a vr and a vq =wr. But as your r changes, you will need to increase your vq if your are to keep your w constant! This is the source of the acceleration that we call the Coriolis acceleration.

An alternative notation that helps us with error checking is:

(dr/dt) = vr = r'

(dvr/dt) = r''

(dq/dt) = w = q'

(dw/dt) = a = q'' .

Using these, we get for the velocity:

v  =  r'r + rq'q .

Note that each term must contain a distance (r) and a time derivative (one ').

Using these, we get for the acceleration:

a  =  (r''- rq'2)r + (2r'q' + rq'')q .

Note that each term must contain a distance (r) and two time derivatives (two ').

We'll use this notation as we look at motion in three dimensions.

Homework Problem #15: Find the jerk (da/dt) in polar coordinates.

4. Time Derivative of a vector

a) In rectangular coordinates: A = Axx + Ayy .

dA/dt  =  (dAx/dt)x + (dAy/dt)y  =  Ax'x + Ay'y ,

since all the rectangular unit vectors are constant and so dx/dt = 0, etc.

b) In polar coordinates: A = Ar r + Aq q .

dA/dt  =  (dAr/dt)r + Ar(dr/dt) + (dAq/dt)q + Aq(dq /dt)  =  Ar'r + Arq'q + Aq'q - Aq q'r

=  (Ar' - Aq q')r + (Aq' + Arq')q .

Example:

Let’s use A = v (since we already know what dv/dt = a is in polar coordinates):

Ar = vr = r’,  and  Aq = vq =  rq' , so  dA/dt  =  (Ar' - Aq q')r + (Aq' + Arq')q   becomes  a  =  dv/dt  =  ( d[r’]/dt – [rq'](q') ) r + ( d[rq']/dt + [r’](q') )q   =   [  (r''- rq'2)r + (2r'q' + rq'')q .