Uniform Motion on a Circular Path

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Homework Statement


95] A particle P travels with a constant speed on a circle of radius r=3.00m and completes noe revolution in 20.0s. The particle passes through O at time t=0. State the following vectors in magnitude-angle notation(angle relative to the positive direction of x). With respect to O, find the particle's position vector at the times t of
a) 5.00s
h] next, find the acceleration at the beginning and end of the interal from 5s to 10s.

Homework Equations



T= (2r*pi)/v
S= theta/r

The Attempt at a Solution


It's quite embarrassing that I am in university and having difficulties with such a simple problem, but it's been 8 months since I have worked on physics =[
Anyways, I know that velocity is not constant, but speed is; furthermore, the magnitude of the acceleration should also be constant.

Using this info, I found the velocity easily by using V=d/t, but am having trouble finding the angles. I used the distance that I got from that equation and subbed it into the S=r*theta equation, but I keep getting the wrong answer for a) (I got about 75degrees, the answer key gets 45degrees). I am not sure where I'm going wrong here...

There are other parts to the equation as well, but they are all similar and I can do them just fine if I can find the correct angles. Also, I am not sure how to find the angle on the acceleration at all, so any starting hints would be appreciated :)

Thanks again for your help for a terrible student =3
 
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Answers and Replies

  • #2
Redbelly98
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The particle passes through the following vectors in magnitude-angle notation (angle relative to the positive direction of x).

It seems that there is some information missing from your post?
 
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It seems that there is some information missing from your post?

Wow, I didn`t catch that...
Thanks a lot, I edited my post already.

Still stuck on the question though, I tried for another 5 minutes then gave up and went on to Calculus :p
 
  • #4
Redbelly98
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Where is point O located?
 
  • #5
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In hindsight, I should`ve attached a diagram for this situation.

Basically, imagine a set of coordinate axis (X and Y coordinates). The circle is centered right at the origin, and point O is at the origin.

Sorry for the unclear description.
 
  • #6
Redbelly98
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So where is the particle at time t=0? If it travels around the circle, it can never be at the origin O as you indicated???

edit:
... The particle passes through O at time t=0.
 
  • #7
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So where is the particle at time t=0? If it travels around the circle, it can never be at the origin O as you indicated???

edit:

I have no idea what I`m talking about at the moment. I guess saying that the circle was centered at the origin made no sense; the bottom of the circle is centered at the origin. Here`s an attached diagram, and once again, sorry for the trouble.
 

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  • #8
Redbelly98
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Okay, that helps show what's going on.

A complete revolution takes 20.0 s. So, how far around the circle does the particle go in 5.00 s?
 
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Okay, that helps show what's going on.

A complete revolution takes 20.0 s. So, how far around the circle does the particle go in 5.00 s?

D= Vt, which works out to be 4.71m or something, I don't remember exactly.
How do I find the angle though? I keep getting the incorrect answer by using S=r*theta formula?
 
  • #10
Redbelly98
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You don't need S=r*theta.

5.00 s is ____% of 20.0 s?
 
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Redbelly98
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Yes.

So in 5.00 s the particle goes 25% (that's one quarter) of a full revolution.

If it starts at O, one quarter of a revolution puts the particle where? (Do they say if it's going clockwise or counter-clockwise?)
 
  • #13
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Yes.

So in 5.00 s the particle goes 25% (that's one quarter) of a full revolution.

If it starts at O, one quarter of a revolution puts the particle where? (Do they say if it's going clockwise or counter-clockwise?)

I would assume that it's going CCW, since there is an arrow indicating the direction (I just didn't draw it). So for the velocity and acceleration, I would also just find the position vector's angle and it would remain the same for V and A?
 
  • #14
Redbelly98
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The angle of the velocity and acceleration vectors will be different than the angle of the position vector.

Have you solved/found the correct position vector at 5.00 s?
 
  • #15
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The angle of the velocity and acceleration vectors will be different than the angle of the position vector.

Have you solved/found the correct position vector at 5.00 s?

Yes, I got to the part where it asked me to find the displacement from 5s-10s. I got the right angle (or at least understand how that works) at 135 degrees, but I can't get the right displacement, and thus, I cannot get the right velocity or acceleration either.

If I want displacement, don't I just take the positions at 10s and 5s and subtract them from each other? If I want velocity, don't I just use v2-v1 with the negative of v1 and then divide by the total time?

I have no idea how to find acceleration though.
 
  • #16
Redbelly98
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Yes, I got to the part where it asked me to find the displacement from 5s-10s. I got the right angle (or at least understand how that works) at 135 degrees, but I can't get the right displacement, and thus, I cannot get the right velocity or acceleration either. If I want displacement, don't I just take the positions at 10s and 5s and subtract them from each other?
When I read your original post, it asks for the displacement at 5.0 s, with respect to O. But here, you say you are asked to find the displacement from 5s to 10s.
Since those are different questions, can you clarify what is being asked?

You're correct about the displacement from 5s to 10s, assuming you subtract the 5s position from the 10s position and not the other way around.

If I want velocity, don't I just use v2-v1 with the negative of v1 and then divide by the total time?
That will give you the average acceleration over some time interval. (You can quickly check if an idea might be wrong by looking at the units. What you suggest is dividing a velocity by a time, giving m/s/s = m/s2 = acceleration units, not a velocity.)

If you want the velocity, figure out where the particle's position is. You know it moves around the circle, so you can look at the figure to figure out the direction it's moving in at any position.

I have no idea how to find acceleration though.
Look at where your physics textbook talks about uniform circular motion. It tells you the magnitude and direction of acceleration for a particle moving in a circle at constant speed.

p.s. signing off for the day, good luck!
 
  • #17
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When I read your original post, it asks for the displacement at 5.0 s, with respect to O. But here, you say you are asked to find the displacement from 5s to 10s.
Since those are different questions, can you clarify what is being asked?

You're correct about the displacement from 5s to 10s, assuming you subtract the 5s position from the 10s position and not the other way around.


That will give you the average acceleration over some time interval. (You can quickly check if an idea might be wrong by looking at the units. What you suggest is dividing a velocity by a time, giving m/s/s = m/s2 = acceleration units, not a velocity.)

If you want the velocity, figure out where the particle's position is. You know it moves around the circle, so you can look at the figure to figure out the direction it's moving in at any position.


Look at where your physics textbook talks about uniform circular motion. It tells you the magnitude and direction of acceleration for a particle moving in a circle at constant speed.

p.s. signing off for the day, good luck!

Thanks for all of your help, I think I've got most of the answers correct. Sorry for not being clear in the question, but there were like 6 different parts to it, so I only posted the questions that seemed unclear to me (the ones that I spent like 3 hours on to no avail).
I had to find position at 5s,7.5s, and 10s, then I had to find velocity from 5-10s, then acceleration at those points. It was confusing since I got mixed up with the answer key (still getting the answer to my displacement wrong).

Anyways, thanks again :D
 

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