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).
2. Velocity
a) Rectangular Coordinates:
v = dr/dt = d[xx + yy]/dt = (dx/dt)x + (dy/dt)y = vxx + vyy .
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) x + sin(q) y]/dt =
-sin(q) (dq/dt) x + cos(q) (dq/dt) y . We now define w =dq/dt, and we note that
q = -sin(q) x + cos(q) y. This gives the result:
dr/dt = (dr/dq)(dq/dt) = wq .
Note in particular that (dr/dq) = q . Using vr = dr/dt, we have:
v = vrr + rwq = vrr + vq q ,
where vq = rw . 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 ra as the normal tangential acceleration. (4) But what is the 2vrw 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'r + 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 .