Find a vector to a particle from a rotated reference frame

[elmers2424]

Homework Statement

Consider a particle in space whose position vector from the origin of reference frame i is
given by the expression r(Oi to P) = 37t(Ii + Ji + Ki). The distance vector from the origin of
another reference frame (m) to the origin of the I reference frame is given by
r(Om to Oi) = e^(-2t)Ii. The i and m frames are initially aligned at time equal zero, but over time the m reference frame wobbles or oscillates about the Ki axis. The rotation angle (θ) between the two reference frames oscillates according to θ = sin(13t). Determine the distance vector to the particle from an observer on reference frame m.

Homework Equations

No equations really...getting the correct components is what I'm after.

The Attempt at a Solution

I first setup reference frame i and placed point P somewhere in positive Ii, Ji, Ki space. Next, I drew another reference frame (m), such that I am and Ii are in the same line of action. All other planes are parallel. I went ahead and put the distance vector r(Om to Oi) on the I am axis (directionally equivalent to the Ii axis). Then, I drew the m reference frame again. This time rotated by some arbitrary angle to get an idea of what it looks like. From the origin in m, I drew a line to P. Now, I need to find the Im, Jm, and Km components. I know that part of my I am component will have the magnitude of e^(-2t) included. For this component I think the full term would be (e^(-2t) + 37tcos(sin(13t))Im. My other thought is because I can use the "tip to tail" method with vectors, all I need to show is (e^(-2t)Im + 37t(Ii)). However, now I don't know how to translate Ii to the I am reference frame. Then, I just don't know what to do to try and determine the Jm and Km components to P. I'm not very good at visualizing 3D space. Any help would be appreciated!
Thank you

 

[anathema]

for your question! It seems like you have a good start on visualizing the problem and setting up the reference frames. To find the components of the distance vector to the particle from an observer on reference frame m, you will need to use some vector algebra and trigonometry.

First, let's consider the rotation of the m reference frame about the Ki axis. We can represent this rotation as a rotation matrix, which will rotate vectors in the m reference frame to vectors in the i reference frame. This rotation matrix can be written as:

R = [cosθ -sinθ 0; sinθ cosθ 0; 0 0 1]

Remember that the rotation angle, θ, is given by θ = sin(13t). So, at any given time t, we can determine the rotation matrix R.

Next, we want to find the position vector of the particle from the origin of reference frame m. This can be written as:

r(Om to P) = r(Om to Oi) + r(Oi to P)

Using the information given in the problem, we can write this as:

r(Om to P) = e^(-2t)Ii + 37t(Ii + Ji + Ki)

Now, we need to rotate this vector using the rotation matrix R to get it in the m reference frame. This can be written as:

r(Om to P) = R * (e^(-2t)Ii + 37t(Ii + Ji + Ki))

Using matrix multiplication, we can simplify this to:

r(Om to P) = (e^(-2t)cosθ + 37tcosθ)Ii + (e^(-2t)sinθ + 37tsinθ)Ji + (e^(-2t) + 37t)Ki

Now, we can see that our distance vector to the particle from an observer on reference frame m has components in the Im, Jm, and Km directions. To find the magnitude and direction of this vector, we can use the Pythagorean theorem and trigonometry.

The magnitude of the distance vector can be found using:

| r(Om to P) | = √((e^(-2t)cosθ + 37tcosθ)^2 + (e^(-2t)sinθ + 37tsinθ)^2 + (e^(-2