Measuring parallel velocity in rotating frame.

[peripatein]

Homework Statement

At t=0 a police officer is located at (0,R) on a circular platform whose radius is R and which rotates around the z axis with constant angular velocity ω. The officer's velocity at that point in time is (ωR ,0). At that time, a bird leaves the center of the platform along the x-axis with constant velocity (v₀,0). The officer is equipped with a laser velocity-meter which measures the velocity parallel to the beam. The officer aims the device at the bird. What would be the velocity of the bird as detected by the officer?

Homework Equations

The Attempt at a Solution

The officer's velocity should be (I think):
V_officer = ωR(-sin(ωt),cos(ωt)). However, I am not sure what the bird's velocity would be. In the lab's reference frame the bird would obviously be moving in a straight line. Moreover, if in the rotating reference frame its velocity is constant then wouldn't its velocity simply be: v₀(-sin(ωt),cos(ωt))? I am quite sure this is wrong, but I am making an effort. I'd appreciate some help with this.

 

[voko]

Is your velocity formula for the officer compatible with the initial conditions?

Regarding the rest of the problem, both velocities are vectors. What is the bird's velocity relative to the officer?

 

[peripatein]

Right, the officer's velocity should be:
ωR(cos(ωt),sin(ωt))
Relative to the officer, the bird is flying with additional v₀ along the x-axis, isn't it?

 

[voko]

Where should the officer be when $\omega t = \pi / 2$? Where should the velocity be directed? Is that compatible with your formula?

The bird is flying along the x-axis in the lab frame. The officer is measuring velocity in the officer's frame. How can you reconcile the two?

 

[peripatein]

Let's focus on the bird for a second. Silly me, for some peculiar reason I figured it would be moving along the x-axis in the rotating frame instead of flying in the lab. In any case, the reconciliation should be via:
v_rotating=v_lab - ωXr (all these are vectors of course).
After substituting the appropriate relations for the unit vectors as well, I obtained:
v_bird=(v0cos(ωt)-v0ωtsin(ωt),-v0sin(ωt)-v0ωtcos(ωt)) in the rotating frame.
Does that seem correct to you?

 

[voko]

No, that does not seem correct. Especially because the bird's velocity is growing unbounded over time.

Is there any reason why you have to convert the bird's velocity to the rotating frame?

 

[peripatein]

Would it be better to leave it as v0 in the lab's RF and convert the officer's velocity to the lab's RF?

 

[voko]

You don't have to convert the officer's velocity to the lab frame. You only have to get it right, as I suggested in #2 and #3.

 

[peripatein]

The platform rotates clockwise (not counter-clockwise). Hence, at y=R and t=0, the angle should be zero I think, and not pi/2. Therefore, the officer's velocity should be:
ωR(cos(ωt),-sin(ωt))
Is this correct?

 

[voko]

That looks good to me.

 

[peripatein]

Okay, now is this the officer's velocity in the lab reference frame?
Also, how do I proceed? Do I now need to subtract this velocity from the bird's and multiply by the cosine of the angle between the two velocities (to obtain the velocity parallel to the officer's)?

 

[voko]

Why do you need a "velocity parallel to the officer's"? The officer's device measures "the velocity parallel to the beam". What is "the beam" here?

 

[peripatein]

Is the beam tantamount to the bird?

 

[peripatein]

Would you be so kind as to at least list the steps necessary for solving this problem, having determined the officer's velocity?

 

[peripatein]

I know the final answer, but am not sure how to obtain it. The final answer is:
{v₀²t-v₀R[sin(ωt)+ωtcos(ωt)]}/√[R²+(v₀t)²-2v₀Rtsin(ωt)]
I don't have much time. Am doing this for an exam I shall have to take within a couple of hours and after the exam it will be far less relevant. If you could guide me through by listing all the necessary steps I'd appreciate it.

 

[voko]

The "beam" is the line of sight from the officer to the bird. You need to obtain the projection of the bird's relative velocity (relative to the officer's velocity) onto the beam.

 

[peripatein]

Do I need to find the vector connecting the officer and the bird then? And then project the relative velocity onto that vector?

 

[peripatein]

The bird's position is given by v0t, whereas the officer's could be obtained by integrating his velocity vector, couldn't it?

 

[voko]

Yes that is correct. Even though you should have been able to guess the officer's position from the data given.

 

[peripatein]

Alright, so upon finding the vector connecting the two, do I simply multiply the bird's velocity vector by the cosine of the angle between the former vector and the latter (that is, the angle between the beam and the bird's velocity)?

 

[voko]

You would need to know the angle for that, but that is doable. A more direct approach would use the properties of the scalar (dot) product.

 

[peripatein]

That is what I meant by the cosine of the angle. By the properties of the dot product, the cosine would be equal to the scalar product of the two vectors, divided by the product of their norms. Am I missing something else to solve this?

 

[voko]

That looks good.

 

[peripatein]

Great, thank you!

 

[peripatein]

Last time we discussed this I had the impression I had everything I needed in order to solve this problem. Upon returning to it, it's become apparent that I seemingly do not. The relative velocity should be:
(v0-wRcos(wt),sin(wt)) whereas the vector connecting the bird and the officer should be:
(v0t-Rsin(wt),-cos(wt)). Do you agree with that? However, projecting the relative velocity vector onto the vector connecting the two objects does not yield the correct answer. Any advice? What am I doing wrong?

 

[voko]

If at t = 0, the officer's velocity was (ωR, 0), then the velocity as a function of time can only be ωR(cos(ωt), ±sin(ωt)). Note the ± sign. Integrating that, the position must be R(sin(ωt), - ±cos(ωt)), and, because we are given the position at t = 0 as (0,R), the ± sign must be the minus sign. So the velocity is ωR(cos(ωt), -sin(ωt)), and the position is R(sin(ωt), cos(ωt)). The relative velocity is the bird's velocity minus the officer's velocity, which is (v - ωR cos(ωt), ωR sin(ωt)). The relative displacement is the bird's position minus the officer's position, so it is (vt - R sin(ωt), -R cos(ωt)). That gives me the answer you specified in #15.

 

[peripatein]

Please tell me, voko, did you obtain that answer through the following?:
(v - ωR cos(ωt), ωR sin(ωt))*cosine of angle between the relative displacement and relative velocity=
(v - ωR cos(ωt), ωR sin(ωt))*{(v - ωR cos(ωt), ωR sin(ωt))*(vt - R sin(ωt), -R cos(ωt))/sqrt(|RD||RV|)
where RD=(vt - R sin(ωt), -R cos(ωt)) and RV=(v - ωR cos(ωt), ωR sin(ωt))

 

[voko]

I projected the relative velocity vector onto the unit direction vector using the scalar/dot product.

That seems to be what you are trying to do as well, but I cannot makes sense of your final formula, it has one multiplication too many, and there is a mismatched bracket.

 

[peripatein]

Did you then simply carry out the following?:
(v - ωR cos(ωt), ωR sin(ωt))*(vt - R sin(ωt), -R cos(ωt))/sqrt(|RD|)

 

[peripatein]

If I am mistaken again, it would be very helpful if you wrote down *your* final formula.

 

[voko]

Yes, except it is /|RD|, not /sqrt(|RD|).