Dealing with Rotating Coordinate Systems Physics 321. (Eq.1)

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1 Dealing with Rotating Coordinate Systems Physics 321 The treatment of rotating coordinate frames can be very confusing because there are two different sets of aes, and one set of aes is not constant in time. This handout is an attempt to clarify the problems of transforming position, velocity, and acceleration in a fied frame of reference to a rotating frame of reference. Because we are dealing with rotations, it is useful to write the vectors as column vectors so that we can apply rotation matrices to them. Usually we just write: r = y = + yy +, but because the basis vectors are implied and not eplicitly written, we are going to write the same thing as a product (the dot product) of a basis set and the components in that basis. Thus: r = ( y ) y = + yy +, Let s take two sets of coordinate aes that have the same origin. There eists a linear transformation that relates one set of basis vectors to the other set of basis vectors. Since the lengths of the basis vectors (they re all unit vectors) do not change, we can think of this transformation as a rotation, or actually three rotations around different aes in general. Thus we can write: ( y ) = ( y )R or ( y ) = ( y )R 1 (Eq.1) where the primes indicate quantities in the rotating coordinate system and the unprimed quantities represent quantities in the fied coordinate frame. Note that the rotation matri follows the vector because we are rotating a row vector. We ll later see how to write R in terms of rotation angles. A position vector is just an object defined by two points in space. It can be described without a coordinate system of any kind, or it can be written in terms of components. Fig. 1 illustrates this idea. Figure 1. A vector in terms of two different sets of basis vectors/

2 Since the vector is the same object in either coordinate system, we can write: r = r ( y ) y = ( y ) y We use a red equals sign just to remind us that we are using different sets of unit vectors on each side of the equation, so we can t equate components of the vectors! Now we can do some manipulating with the rotation matri. Noting that any matri times its inverse is the identity: Because of Eq.1, we may write: ( y ) y = ( y )R R y y = R y or y = R y (Eq. 2) So now we can transform a position vector both components and unit vectors from the fied to the rotating frame, as long as we know the rotation matri that relates the two frames to each other. Now let s look at velocities: ( y ) y = y d dt r = d dt r y + ( y ) y The only hard part here is taking the derivatives of the primed unit vectors. d dt ( y ) = d ( y )R = ( y )R dt y = ( y )R R (Eq.3) Armed with these, we can write the equations that transform the components of velocity from one coordinate frame to the other. First, let us put everything in terms of the unprimed basis vectors.

3 ( y ) y = ( y )R y + ( y )R y y = R y + R y (Eq.4) This epression then gives the components of velocity in the fied system in terms of the rotation matri and the components of position and velocity in the unprimed system. We can similarly find the components of the velocity in the primed system in terms of the components of position and velocity in the unprimed system. ( y )R y = ( y )R R y + ( y )R R y y = R y R RR y (Eq.5) A more familiar way of writing these equations can be obtained by going back to the line before Eq. 3 and writing the result in mied unit vectors: ( y ) y = y y + ( y ) y ( y ) y = ( y )R R y + ( y ) y ( y )v = ( y )R R r + ( y )v ( y )v = ( y )v + R R r ( y )v = ( y )[v + ω r ] Here, ω is the angular velocity of the rotating system as measured in the fied frame, but whose components are measured in the rotating frame, not in the fied frame. That is: ω = R ω where ω is the angular velocity of the rotating frame as measured in the fied frame with the unit vectors of the fied frame.

4 Therefore, as long as we remember that everything on the left-hand side is epressed in terms of the unit vectors of the fied frame, and everything on the right-hand side is epressed in terms of the rotating frame (hence the red equals sign), we have: v = v + ω r (Eq.6) The term with the rotation matrices can also be written in terms of the unprimed vectors as follows: ( y ) y = y y + ( y ) y ( y ) y = ( y )R R y + ( y ) y ( y ) y = ( y )RR R R y + ( y ) y ( y )v = ( y )R R r + ( y )v ( y )v = ( y )v R R r ( y )v = ( y )[v + ω r ] v = v ω r In these epressions, we have made use of the identities (Eq.7) R R r = ω r and R R r = ω r which we will prove later. We ll also discuss rotation matrices more later. For now we ll just give the form for R when the rotation is about the -ais. cos ωt sin ωt R = sin ωt cos ωt, 1 cos ωt sin ωt R = sin ωt cos ωt 1 Now let s try applying these ideas to a sample problem.

5 Eample 1: A frictionless block moves radially outward from the center of a turntable. Let us take the velocity of the block in the lab to be v = v. Find the position and velocity of the block as measured by someone on the turntable. Take the angular speed of the turntable to be ω. First, let s recapitulate what we know: ω = ω, v = v, r = v t Eq. 7 gives: ( y )v = ( y )[v ω r ] = v ωv t = v ωv ty For position, we have: v t v t ( y )r = ( y ) = ( y )R cos ωt sin ωt v t cos ωt r = sin ωt cos ωt = v t sin ωt 1 And for velocity: ( y )v = ( y )R v ωt cos ωt sin ωt v ( y )v = ( y ) sin ωt cos ωt v ωt 1 cos ωt ωtsin ωt v = v sin ωt + ωt cos ωt Note that the position and velocity are related by v = dr /dt, as epected. v

6 General Results Equations 6 and 7 apply specifically to velocities, but in general we can apply the same techniques to other vector quantities as well. To do this we start with the epression for some vector Q that relates its form in the primed and unprimed coordinates: ( y )Q = ( y )Q We then put that epression for Q in the equation ( y )Q = ( y ) Q + ω Q. So if Q = r, and ( y ) y = ( y ) y ( y )r = ( y )r + ω r ( y )v = ( y )[v + ω r ] as we demonstrated before. If, however, Q = v, the process is a little more complicated. To write the epression ( y )Q = ( y )Q, we have the following: ( y )v = ( y )[v + ω r ]. That is, in the general form above, we must replace Q with v + ω r, not just v. ( y )Q = ( y ) Q + ω Q ( y )v = ( y ) d (v + ω r ) + ω (v + ω r ) dt ( y )v = ( y )v + ω r + ω r + ω v + ω (ω r ) ( y )a = ( y )a + ω r + 2ω v + ω (ω r ) a = a + ω r + 2ω v + ω (ω r ) (Eq. 8)

7 Similarly, it can be shown that: ( y )Q = ( y ) Q ω Q ( y )v = ( y ) d (v ω r ) ω (v ω r ) dt ( y )v = ( y )v ω r ω r ω v + ω (ω r ) ( y )a = ( y )a ω r 2ω v + ω (ω r ) a = a ω r 2ω v + ω (ω r ) (Eq. 9) What we usually want to know in the end is the effective force that we feel in our rotating coordinate system. That typically involves multiplying every term in Eq. 8 by m and solving for F = ma. Note that F = ma is just the total true force (including gravity, normal forces, etc.). In Eq. 8, this force is epressed in the ( y ) basis set, but if we know that force in the primed basis set, then everything is in the primed basis. That is: ( y )[ma ] = ( y )ma + mω r + 2mω v + mω (ω r ) F = F + mω r + 2mω v + mω (ω r ) F = F mω r 2mω v mω (ω r ) F = F + r mω + 2mv ω + m (ω r ) ω (Eq.1) F = F + F + F + F Eample 2: Returning to Eample 1, we wish now to calculate the effective acceleration eperienced by the block on the turntable. ( y )a = ( y )a ω r 2ω v + ω (ω r ) ( y )a = ( y )[ 2ω v + ω (ω r )] ( y )a = 2ωv + ω v t ( ) = 2ωv y ω v t ω v t ( y )a = ( y ) 2ωv But note that we have a strange epression here. This is the acceleration in the turntable frame, but epressed in terms of the rest frame basis vectors. So we need to do a rotation before we re finished.

8 ω v t cos ωt sin ωt ω v t ( y )a = ( y )R 2ωv = ( y ) sin ωt cos ωt 2ωv 1 ω v t cos ωt 2ωv sin ωt a = +ω v t sin ωt 2ωv cos ωt If we multiply this by mass, this would represent the force that would be necessary to apply to the block for it to move on a non-rotating (frictionless) floor to move the same way that it does on the turntable. Notes on Rotation Matrices: To begin with, we wish to find the rotating unit vectors in terms of the fied unit vectors. If the rotation is about the -ais, we have, with θ = ωt: = cos ωt + sin ωt y y = sin ωt + cos ωt y = Putting this in matri form, we have cos ωt sin ωt y = sin ωt cos ωt y. 1 Figure 1. Rotation of basis vectors about the ais. To relate this to the rotation matri we defined, it is useful to prove an identity. To do this, we use Eq.1: ( y ) = ( y )R Now we take the transpose of both sides, noting that (AB) = B A : But we already saw that y = R y y = R y.

9 So we can conclude that R = R. Now we can write the rotation matri for rotations around the ais: cos ωt sin ωt R = sin ωt cos ωt Similarly, we can show: 1 1 R = cos ωt sin ωt sin ωt cos ωt cos ωt sin ωt R y = 1 sin ωt cos ωt Now we have established the form of rotation matrices about the three aes, it is useful to think about infinitesimal rotations. For eample, let s take a rotation by a small angle about the ais. cos ω sin ω 1 ω R (ω) = sin ω cos ω ω We can generalie this to a sequence of three infinitesimal rotations, one around each ais: 1 ω 1 ω 1 R(ω, ω, ω ) ω ω 1 ω 1 ω 1 1 ω ω = ω 1 ω ω ω 1 If the rotations are infinitesimal, the order of rotation does not matter. We can then write, using 1 to denote the identity matri: Rω, ω, ω 1 ω ω ω ω ω ω Thus, any infinitesimal rotation can be considered to be a rotation by an angular velocity ω about an ais described by ω.

10 Now, let s look at how components of vectors in the fied and rotating frames are related at two times, t and t = t +. y = R y y = R y R = R + R = R R We can write this last line because we can think of rotation 2 as rotation 1 followed by an infinitesimal rotation about ω. (The first rotation is the one nearest the vector upon which it operates, so the first rotation is to the right of the second rotation in the epression.) We can now solve for the infinitesimal rotation: R R = R + R now by R R = 1 + R R ω R ω 1 ω = R R = ω ω ω ω ω R ω 1 ω ω ω y r = ω ω y = ω ω = ω r ω ω ω y ω We used this result to Prove Eq. 7. The question then is, in which coordinate system is ω measured? Since we rotate both the unit vectors and the components of matrices, this can get confusing. But we can answer this question by looking at the equation: R r = R R r. We start in the primed system, but then we rotate by R, which, ignoring the infinitesimal rotation R, puts us in the unprimed system. Hence, when R acts on the system, it is in the fied coordinate system. That means we need to put ω in the unprimed system when we use the equation v = v ω r

11 We can now do the same thing, but use the inverse transformation: y = R y y = R y R = R + R = R R We again now solve for the infinitesimal rotation. This time, we are solving for the components of the infinitesimal inverse rotation, so we are solving for the components of ω. R R = R + R now by R R 1 R = 1 + R R +ω ω = R R = ω +ω +ω ω What we really want this time, however, is R R = R R. To find that, we can make use of the relationship: Then: 1 R R =, 1 = = R R + R R R R = R R R 1 This time we have: R 1 ω ω = R R = ω ω ω ω ω ω r = ω ω ω ω ω ω y = ω ω = ω r ω y ω y R = R R We start in the unprimed system, but then we rotate by R, which puts us in the primed system. Hence when R acts on the system, it is in the rotating coordinate system. Thus, we need to put ω in the primed system when we use the equation v = v + ω r.