Tangential Velocity (maybe)

  • #1
Old Man Scho
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OP warned for not providing an attempt at a solution
TL;DR Summary
Can the tangential velocity of B be found from the information given?
two bodies.jpg
 
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  • #2
Welcome to PF.

Yes.
I assume, tangential to a circle about the origin ?
The position of point B is the sum of the two rotating vectors.
The angle between the origin and B changes with time.
The rate of change of that angle is the angular velocity = tangential velocity.
 
  • #3
Old Man Scho said:
Can the tangential velocity of B be found from the information given?
Not quite. Additional needed information is the position of points A and B at time ##t=0##. Then one can find the velocity of B at any later time ##t##.
 
  • #4
kuruman said:
Additional needed information is the position of points A and B at time ##t=0##.
Do you mean the position in terms of a point on the circle as a degree, like 90°?
 
  • #5
Old Man Scho said:
Do you mean the position in terms of a point on the circle as a degree, like 90°?
Yes. It could be an "o'clock" position for each of the points A and B. In the picture you posted A is at 3:00 and B is at about 1:30, but it doesn't have to be that.
 
  • #6
Baluncore said:
I assume, tangential to a circle about the origin ?
So does the body B already have the linear velocity of body A's? I'm assuming that it must, right? But it's not as simple as adding the two linear velocities together is it?
 
  • #7
kuruman said:
Yes. It could be an "o'clock" position for each of the points A and B. In the picture you posted A is at 3:00 and B is at about 1:30, but it doesn't have to be that.
Ok, then can we solve this using 3 and 1:30?
 
  • #8
Here are some arbitrary values:

A ω=.5 radians/sec
B ω = 2 radians/sec
r1 = 10m
r2 = 3m

v = r*ω

Would I calculate each linear velocity to start with?
 
  • #9
Try to do this using symbols, not numbers. It will be easier for us to check your work. Use LaTeX for writing equations. If you don't know how, click on the link "LaTeX Guide", lower left above "Attach fiiles."
Let the angular velocities be ##\omega_A## and ##\omega_B## and the positions be ##r_{\!A}## and ##r_{\!B}.##
Old Man Scho said:
Would I calculate each linear velocity to start with?
Yes, but to do that you first need to find ##r_{\!A}## and ##r_{\!B}## as functions of time.
 
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  • #10
Baluncore said:
Welcome to PF.

Yes.
I assume, tangential to a circle about the origin ?
The position of point B is the sum of the two rotating vectors.
The angle between the origin and B changes with time.
The rate of change of that angle is the angular velocity = tangential velocity.
Please tell me why B will start rotating?
 
  • #11
titasdasplus said:
Please tell me why B will start rotating?
The OP diagram shows B rotating about A, while A rotates about the origin.
 
  • #12
titasdasplus said:
Please tell me why B will start rotating?
Because ##\omega_B## is assumed to be non zero.
 
  • #13
kuruman said:
Because ##\omega_B## is assumed to be non zero.
Please give more description.
 
  • #14
kuruman said:
Try to do this using symbols, not numbers.
I will do my best.

kuruman said:
to do that you first need to find r_{\!A} and r_{\!B} as functions of time.
Ok, I'm not sure how to find that. I thought r is a function of length.
 
  • #15
##r## is the length of a position vector from a fixed origin. Look at your drawing. Both ##r_1## and ##r_2## depend on time as point A goes around the large circle whose center is fixed and point B goes around the smaller center whose center is point A and not fixed.
Old Man Scho said:
I will do my best.
I wouldn't expect anything less than that. Please post it so that we can help you move along. Begin by finding the tangential velocity of point A.
 
  • #16
titasdasplus said:
Please give more description.
What additional description do you require?
 
  • #17
kuruman said:
##r## is the length of a position vector from a fixed origin. Look at your drawing. Both ##r_1## and ##r_2## depend on time as point A goes around the large circle whose center is fixed and point B goes around the smaller center whose center is point A and not fixed.

I wouldn't expect anything less than that. Please post it so that we can help you move along. Begin by finding the tangential velocity of point A.
Using the numbers I posted:

vt = ω r

vt = .5 x 10 = 5 m/s
 
  • #18
Velocity is a vector and has both magnitude and direction. The magnitude (also known as speed) is constant but its direction changes depending on the time you look at it. How can you express the velocity as a function of time? Hint: Write the position of point A as a function of time and take the time derivative.
 
  • #19
kuruman said:
How can you express the velocity as a function of time?
radians per second, revolutions per minute, miles per hour?

kuruman said:
Write the position of point A as a function of time and take the time derivative.
I'm not sure how to take a time derivative.
 
  • #20
Old Man Scho said:
radians per second, revolutions per minute, miles per hour?
These are units. I am asking you to write an algebraic expression for the velocity at any time ##t##. In other words, I give you a time, say 2 seconds and you give me speed and the direction at that time.
Old Man Scho said:
I'm not sure how to take a time derivative.
What is your level of education in math and physics?
It looks like you may not have the necessary background to complete this task. Where did this problem come from and why do you need the answer?
 
  • #21
I have never studied calculus if this is what you mean. Going forward I am attempting to learn what is necessary to solve this. I created the problem and I need the answer because I want to know. Understand?
 
  • #22
Have you studied vectors? Will you be able to understand the answer if it is given to you?
 
  • #23
If you do not know how to solve this problem then say so and perhaps someone else can actually help . I asked for the velocity of B and it can be derived from the information without any vectors being necessary.
 
  • #24
Old Man Scho said:
I asked for the velocity of B and it can be derived from the information without any vectors being necessary.
If you insist. So let's forget about vectors and use a drawing. Shown below is the trajectory of point B using your numbers and assuming that A starts at the 3 o' clock position and B at the 2 o'clock position. The velocity of point B is tangent to the trajectory at the point of your choice.

Orbit.jpg
 
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  • #25
  • #26
But will you be able to solve the general problem? Shown below is a trajectory with everything the same except that ##\omega_B##=11 rad/s. Could be a piece of jewelry.

Orbit_2.jpg
 
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  • #27
kuruman said:
If you insist.
You are a saint :smile:

BTW 1: According to my calculations, the problem can be solved "brute force" by using vectors and their derivatives. Although it's not difficult, the final answer as a function of ##~r_1,\omega_1,r_2,\omega_2~## is somewhat messy, and I doubt whether it can be arrived at easily without these mathematical tools.

BTW 2: I suspect that the answer which is given in the reference in post #25 is incorrect.

BTW 3: What software do you use to produce these nice drawings?
 
  • #28
JimWhoKnew said:
BTW 1: According to my calculations, the problem can be solved "brute force" by using vectors and their derivatives. Although it's not difficult, the final answer as a function of ##~r_1,\omega_1,r_2,\omega_2~## is somewhat messy, and I doubt whether it can be arrived at easily without these mathematical tools.
Around 200 CE, Ptolemy came close to describing planetary motion within the geocentric model. He used the idea of epicycles which is essentially what we have here. I suppose a trajectory can be drawn using geometrical consderations. Fifteen centuries after Ptolemy, Isaac Newton invented calculus and used it to describe planetary motion within the heliocentric model.
JimWhoKnew said:
BTW 2: I suspect that the answer which is given in the reference in post #25 is incorrect.
I did not solve that problem but if it "looks fine" to @haruspex, then it is fine.
JimWhoKnew said:
BTW 3: What software do you use to produce these nice drawings?
I use Grapher.app which came bundled with my Mac laptop's operating system.
 
  • #29
kuruman said:
Around 200 CE, Ptolemy came close to describing planetary motion within the geocentric model. He used the idea of epicycles which is essentially what we have here. I suppose a trajectory can be drawn using geometrical consderations.
Thanks for the reminder. From the little I know, calculations by epicycles are not trivial (I didn't say it was impossible, only "I doubt whether it can be arrived at easily without these mathematical tools").

kuruman said:
I did not solve that problem but if it "looks fine" to @haruspex, then it is fine.
A very scientific argument :wink:

Edit: I looked at that other thread once again. The OP was looking for the acceleration and tangential velocity at a single instant, when the radii are aligned (rather than as a function of time which is what I've calculated). If the OP in the present thread is also interested only in a single instant, that should be easier to calculate by trigonometry.

kuruman said:
I use Grapher.app which came bundled with my Mac laptop's operating system.
Thanks
 
Last edited:
  • #30
JimWhoKnew said:
A very scientific argument :wink:
In that thread, I did show that the expression gave the right answer in a couple of special cases. Maybe you would find that more persuasive?
 
  • #31
kuruman said:
What additional description do you require?
Actually I want to know why B will start rotating?
 
  • #32
I assume that vector A rotates about the origin, referenced to the x-axis.
Does vector B rotate relative to vector A, or relative to the x-axis ?

Which tangential velocity is required? Is it the rate of point B about the origin in rad/sec, or distance per second about the origin.
Or is it the instantaneous velocity of point B on the x, y plane ?

titasdasplus said:
Actually I want to know why B will start rotating?
Because it is specified as having an angular frequency.
What alternatives are there?
 
  • #33
haruspex said:
In that thread, I did show that the expression gave the right answer in a couple of special cases. Maybe you would find that more persuasive?
You were right, and my "suspicion" was wrong. I misinterpreted the requirement in the OP (see the "edit" in post #29).
 
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  • #34
kuruman said:
Around 200 CE, Ptolemy came close to describing planetary motion within the geocentric model. He used the idea of epicycles which is essentially what we have here
Well, in this case we're describing the orbit of the Moon...
 

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