Physics 350

Lab 7: Rotational Transforms Sample Solutions

As usual, reset things. But in this case we also want to load a couple of prebuilt Maple routines for linear algebra.

>	restart: with(linalg):

1. Identify the matrix for a rotation through angle φ about the positive z axis. Well, this is just the standard rotation matrix we have had in many places (example on p. 416) where we make sure to leave the z-axis unchanged:

>	R := matrix([[cos(phi),-sin(phi),0],[sin(phi),cos(phi),0],[0,0,1]]);

2. Identify the 3×3 matrix for the transformation which brings the point (1,1,1) to the point (0,0,sqrt(3)) via a counterclockwise rotation about the line from the point (1,–1,0) to the origin (looking toward the origin). The rotation is through angle α, where cos (α) = 1/sqrt(3) . This is the same as rotating a line about the z-axis 45? counter-clockwise followed by some rotation about the x-axis. So first I write my matrix for rotating about the z-axis 45?:

>	M1 := matrix([[cos(Pi/4),-sin(Pi/4),0],[sin(Pi/4),cos(Pi/4),0],[0,0,1]]);

>	test := vector ([1,1,1]);

>	*test_rot := evalm(M1 & test);**

Now I have to figure out how much I have to rotate by to get this rotated vector onto the z-axis, I can use a dot product (since A•B = ABcos(theta) ==> theta = arccos(A•B/AB) ) to figure out the angle between this and the "final" vector we want:

>	final := vector ([0,0,sqrt(3)]);

>	theta := arccos(dotprod(test_rot, final) / dotprod(final,final)); cos_angle := cos(theta); sin_angle := sin(theta);

So the cosine of the angle of rotation is 1/sqrt(3), therefore the rotation matrix for this second rotation about the x-axis is:

>	M2 := matrix([[1,0,0],[0,cos_angle,-sin_angle],[0,sin_angle,cos_angle]]);

So the matrix representing the rotation is the matrix product of these two rotations:

>	*M := evalm(M2 & M1);**

>	*test_rot2 := simplify(evalm(M & test));**

Finally I am asked to have Maple compute the inverse of this matrix:

>	Minv := evalm(1/M);

3. Our next task is to find the matrix that rotates counterclockwise by an angle φ about the line segement from point (1,1,1) to the origin. We are given the hint that we should be thinking of Minv*R*M. In in fact, this makes sense, because what we are doing is rotating the axis of rotation onto the z-axis, then applying a rotation about the z-axis (which corresponds to our original (1,1,1) axis), then inverting the rotation. Therefore a quick bit of Maple should find this rotation matrix:

>	*S := simplify(evalm( Minv & R &* M));**

a. And now we confirm that a rotation of 120? just swaps axes. To do this, I substitute phi = 120? = 2*Pi/3 (radians) into the matrix S and then test the resulting matrix:

>	*S120 := simplify(subs(phi=2Pi/3,matrix(S)));**

>	ihat := vector ([1,0,0]); jhat := vector([0,1,0]); khat := vector([0,0,1]);

>	*inew := evalm( S120 & ihat); jnew := evalm( S120 &* jhat); knew := evalm( S120 &* khat);**

b. Now see where (1,2,4) goes after a rotation of 60? about this axis.

>	S60 := simplify(subs(phi=Pi/3,matrix(S)));

>	*vec124 := vector([1,2,4]); vec124_new := evalm(S60 & vec124);**

c. See where point (1,1,0) goes after a 90? rotation about this axis.

>	S90 := simplify(subs(phi=Pi/2,matrix(S)));

>	*vec110 := vector([1,1,0]); vec110_new := evalm(S90 & vec110);**

Physics 350:
Computational Methods
for the Physical Sciences

Spring Semester 2009

Physics 350: Computational Methods for the Physical Sciences

Spring Semester 2009

Physics 350:
Computational Methods
for the Physical Sciences