Old Man Scho
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OP warned for not providing an attempt at a solution
- TL;DR
- 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##.Old Man Scho said:Can the tangential velocity of B be found from the information given?
Do you mean the position in terms of a point on the circle as a degree, like 90°?kuruman said:Additional needed information is the position of points A and B at time ##t=0##.
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.Old Man Scho said:Do you mean the position in terms of a point on the circle as a degree, like 90°?
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?Baluncore said:I assume, tangential to a circle about the origin ?
Ok, then can we solve this using 3 and 1:30?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.
Yes, but to do that you first need to find ##r_{\!A}## and ##r_{\!B}## as functions of time.Old Man Scho said:Would I calculate each linear velocity to start with?
Please tell me why B will start rotating?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.
The OP diagram shows B rotating about A, while A rotates about the origin.titasdasplus said:Please tell me why B will start rotating?
Because ##\omega_B## is assumed to be non zero.titasdasplus said:Please tell me why B will start rotating?
Please give more description.kuruman said:Because ##\omega_B## is assumed to be non zero.
I will do my best.kuruman said:Try to do this using symbols, not numbers.
Ok, I'm not sure how to find that. I thought r is a function of length.kuruman said:to do that you first need to find r_{\!A} and r_{\!B} as functions of time.
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.Old Man Scho said:I will do my best.
What additional description do you require?titasdasplus said:Please give more description.
Using the numbers I posted: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.
radians per second, revolutions per minute, miles per hour?kuruman said:How can you express the velocity as a function of time?
I'm not sure how to take a time derivative.kuruman said:Write the position of point A as a function of time and take the time derivative.
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:radians per second, revolutions per minute, miles per hour?
What is your level of education in math and physics?Old Man Scho said:I'm not sure how to take a time derivative.
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.Old Man Scho said:I asked for the velocity of B and it can be derived from the information without any vectors being necessary.
A nearly identical problem was posted here previously, so I had already solved mine in the same manner. I was keen to learn of an alternative method and your graph is exactly what I was hoping to see.kuruman said:Have you studied vectors? Will you be able to understand the answer if it is given to you?
You are a saintkuruman said:If you insist.
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 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.
I did not solve that problem but if it "looks fine" to @haruspex, then it is fine.JimWhoKnew said:BTW 2: I suspect that the answer which is given in the reference in post #25 is incorrect.
I use Grapher.app which came bundled with my Mac laptop's operating system.JimWhoKnew said:BTW 3: What software do you use to produce these nice drawings?
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: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.
A very scientific argumentkuruman said:I did not solve that problem but if it "looks fine" to @haruspex, then it is fine.
Thankskuruman said:I use Grapher.app which came bundled with my Mac laptop's operating system.
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?JimWhoKnew said:A very scientific argument![]()