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 90^o to the direction of r:

q = cos(q+90^o) x + sin(q+90^o) y .

We recognize that cos(q+90^o) = -sin(q) and that sin(q+90^o) = +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) = v_xx + v_yy ,

since dx/dt = 0 and dy/dt = 0 , since both x and y are constants. Note also that v_x = dx(t)/dt and v_y = 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 v_r = dr/dt and v_q = rw, we have:

v = v_r r + rwq = v_r r + v_q 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[v_xx + v_yy]/dt = (dv_x/dt)x + (dv_y/dt)y = a_xx + a_yy .

b) Polar Coordinates:

a = dv/dt = d[v_rr + rwq ]/dt = (dv_r/dt)r + v_r(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) = v_r. 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 = (dv_r/dt)r + v_rwq + v_rwq + raq + rw(-w)r = [(dv_r/dt) - w²r]r + [(2v_rw)+ra]q .

(1) We recognize (dv_r/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 2v_rw 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 v_r and a v_q =wr. But as your r changes, you will need to increase your v_q 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) = v_r = r'

(dv_r/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 = A_xx + A_yy .

dA/dt = (dA_x/dt)x + (dA_y/dt)y = A_x'x + A_y'y ,

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

b) In polar coordinates: A = A_r r + A_q q .

dA/dt = (dA_r/dt)r + A_r(dr/dt) + (dA_q/dt)q + A_q(dq /dt) = A_r'r + A_rq'q + A_q'q - A_q q'r

= (A_r' - A_q q')r + (A_q' + A_rq')q .

Example:

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

A_r = v_r = r’, and A_q = v_q = rq' , so dA/dt = (A_r' - A_q q')r + (A_q' + A_rq')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 .

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