Graph of Position Seems to be Wrong

In summary, In part A, the point is observed to have velocity Va relative to coordinate system A. Va is also relative to coordinate system B, which is displaced from system A by distance R. In part B, the point is observed to have velocity Vb relative to coordinate system A and dR/dt is found to be 2wlcos(wt). In part C, the point is observed to have velocity Rx relative to coordinate system B, which does not make sense because the particle is supposed to be motion around a circle.
  • #1
GoCubs12
19
0

Homework Statement



By relative velocity, we mean velocity with respect to a specified coordinate system. (The term velocity, alone, is understood to be relative to the observer’s coordinate system.)

(a) A point is observed to have velocity Va relative to coordinate system A.What is its velocity relative to coordinate system B, which is displaced from system A by distance R? (R can change in time.)

(b) Particles a and b move in opposite directions around a circle with angular speed ω, as shown. At t = 0 they are both at the point r = lˆj, where l is the radius of the circle. Find the velocity of a relative to b.

(c) Sketch the motion of the particle.

Homework Equations



Theta=wt

The Attempt at a Solution



I was able to solve part A and found Va=Vb + dR/dt. I'm pretty confident in this answer.

I was also able to solve part B. I found dR/dt to be 2wlcos(wt). I then plugged this into the answer from part A and got Va=Vb + 2wlcos(wt). I'm pretty confident in this answer as well.

My issue is with part C. To graph the position, I integrated the answer from part B and got Rx=Vbt+2lsin(wt). I believe I did this right, but if you graph it the sin function does not go linearly across the graph as it normally does, but instead gradually increases due to the Vbt term in front of it. This does not seem to make sense in the real world because the particles are going around a circle so they should not be gradually getting farther apart over time? I'm pretty stumped on this. Any input about this would be appreciated.

Thanks for all the help!
 
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  • #2
GoCubs12 said:
My issue is with part C. To graph the position, I integrated the answer from part B and got Rx=Vbt+2lsin(wt). I believe I did this right, but if you graph it the sin function does not go linearly across the graph as it normally does, but instead gradually increases due to the Vbt term in front of it.

If we are assuming particle B's coordinate system places (at all times) particle B at the origin of that system, then particle B's velocity in that system is zero.
 
  • #3
Stephen Tashi said:
If we are assuming particle B's coordinate system places (at all times) particle B at the origin of that system, then particle B's velocity in that system is zero.

So since particle B's position would always be (0,0), it could just be left out of all the equations then? Including in part A?
 
  • #4
GoCubs12 said:
So since particle B's position would always be (0,0), it could just be left out of all the equations then? Including in part A?

No. In part a), I interpret the phrase "coordinate system B, which is displaced from system A by distance R" to mean the origin of coordinate system B is displaced from the origin of coordinate system A by displacement vector R. That does not imply that when we do computations of a particle's velocity that we always assume the particle is at the origin of one of the coordinate systems.

In the special case where a problem asks about something "relative to particle A", it usually means we do use a coordinate system where the particle is at the origin of "its" coordinate system.
 
  • #5
Stephen Tashi said:
No. In part a), I interpret the phrase "coordinate system B, which is displaced from system A by distance R" to mean the origin of coordinate system B is displaced from the origin of coordinate system A by displacement vector R. That does not imply that when we do computations of a particle's velocity that we always assume the particle is at the origin of one of the coordinate systems.

In the special case where a problem asks about something "relative to particle A", it usually means we do use a coordinate system where the particle is at the origin of "its" coordinate system.

Thanks that helps a lot. Also, in part B it is talking about being relative to particle B so the Vb term could be left out since it would imply that it is 0?
 
  • #6
GoCubs12 said:
Thanks that helps a lot. Also, in part B it is talking about being relative to particle B so the Vb term could be left out since it would imply that it is 0?

The way I'd phrase it is that the Vb is zero so the Vb term has no effect. "Left out" implies something about notation, which may or may not get the approval of a grader.
 
  • #7
Stephen Tashi said:
The way I'd phrase it is that the Vb is zero so the Vb term has no effect. "Left out" implies something about notation, which may or may not get the approval of a grader.

Thanks for all the help!
 

1. Why does the graph of position seem to be wrong?

There could be several reasons for this. It could be due to incorrect data input, a faulty graphing tool, or an error in the calculations. It is important to double-check all the data and calculations to ensure accuracy.

2. How can I fix a wrong graph of position?

If the error is due to incorrect data or calculations, fixing them will automatically result in a corrected graph. If the problem lies with the graphing tool, try using a different one or adjusting the settings to see if that resolves the issue.

3. Can a graph of position be wrong even if the data is correct?

Yes, there could be errors in the graphing or plotting process that may result in an incorrect graph even if the data is accurate. It is important to check all the steps and make sure they are done correctly.

4. How can I ensure the accuracy of a graph of position?

To ensure the accuracy of a graph of position, it is important to double-check all the data and calculations, use a reliable graphing tool, and follow the correct procedures for plotting the data. It is also helpful to have a second person review the graph for any potential errors.

5. Is it common for a graph of position to be wrong?

It is not uncommon for a graph of position to have errors, especially if there are many data points or if the calculations are complex. However, with careful attention to detail and accuracy, these errors can be minimized or eliminated.

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