What am I Doing Wrong with this Physics Problem?

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[quote]A particle P travels with constant speed on a circle of radius r=3.00m (Figure) and completes one 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 times t of (a) 5.00 s, (b) 7.50 s, and (c) 10.0 s. (d) for the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

http://img46.imageshack.us/img46/449/physicsu.png
[/quote]

First I know that radius = 3.00m, and time to complete one cycle is 20.0 seconds. So I know that T=2(pi)r/v, so 20.0s=2pi(3.00m)/v, so speed = .94247 m/s

Then acceleration = (v^2)/r, so a = ((.94247m/s)^2)/(3.00m), so a=.29608 m/s

I know using Distance = rate x time, that at a time of 5 seconds, and at a speed of .94247 m/s, the object will be 4.71 meters along the circle

using the arclength formula, s=rθ, 4.71=(3)θ, so the angle is 90 degrees.

So at 90, the position vector is 6 meters in the y direction, and zero meters in the x direction (At the very top of the circle)

before I proceed, is this correct?

So is the position vector <0,6,0>?

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I would try using uniform circular motion equations instead of using linear speed. v=rw and theta=wt, then from there just use some trig to find the displacement from O. And I think the position vector after 5 seconds would be <-3,3,0> (I assume your vectors were in i,j,k notation), and assuming point O sits at point <3,0,0>, or the right-most edge of the circle and travelling towards the positive y-axis.

EDIT:
didn’t see the image before…on your image the position vector would be <3,3,0>

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[quote]jdinatale wrote:

[quote]A particle P travels with constant speed on a circle of radius r=3.00m (Figure) and completes one 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 times t of (a) 5.00 s, (b) 7.50 s, and (c) 10.0 s. (d) for the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

http://img46.imageshack.us/img46/449/physicsu.png
[/quote]

First I know that radius = 3.00m, and time to complete one cycle is 20.0 seconds. So I know that T=2(pi)r/v, so 20.0s=2pi(3.00m)/v, so speed = .94247 m/s

Then acceleration = (v^2)/r, so a = ((.94247m/s)^2)/(3.00m), so a=.29608 m/s

I know using Distance = rate x time, that at a time of 5 seconds, and at a speed of .94247 m/s, the object will be 4.71 meters along the circle

using the arclength formula, s=r�?�¸, 4.71=(3)�?�¸, so the angle is 90 degrees.

So at 90, the position vector is 6 meters in the y direction, and zero meters in the x direction (At the very top of the circle)

before I proceed, is this correct?

So is the position vector <0,6,0>?

-[/quote]

I think you’re right. Then again I haven’t looked at physics / math like this in about 5 years so I could be wrong.

P makes 1 complete rotation around the circle in 20 seconds. Thus, in 5 seconds, it would make 1 quarter of the way around the rotation. 5s/20s = 1/4. So the position vector would be <0,6,0>.

I kind of skipped the whole arc length thing and got to your answer, which makes me think you’re on the right track.

EDIT: Woops… I thought you had started at <3,3,0> when in fact you had started at <0,0,0>. The guy above me would be right at t=5 you would be at <3,3,0>

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You know teachers usually remember students who come to see them w/ legitimate questions when it comes to giving grades.

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[quote]theuofh wrote:
You know teachers usually remember students who come to see them w/ legitimate questions when it comes to giving grades. [/quote]

Definately boosted my GPA quite a bit just by doing this.

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At 5 seconds or 1/4 the way around the circle, wouldn’t you be at <3,3,0> relative to O?

I don’t think you can use Cartesian velocity equations for this. The problem gives you angular velocity (1 rev/20 sec), you should just use that to determine position, for example:

at 5 seconds, (1 rev / 20 sec) = (X rev / 5 secs), now solve for X to get 0.25 revolutions, or 90 degrees.
at 7.5 seconds, (1 rev / 20 sec) = (X rev / 7.5 secs), now solve for X to get 0.375 revolutions, 135 degrees.
etc…

You only need arc length stuff for part d.

Good luck!

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This is a basic rotational motion problem. If you’re having trouble with this then I suggest you read your textbook or ask your teacher.

At 10 s, the position vector is [0,6,0], but this is not magnitude-angle notation. Rewrite is as [6 m, 90 degrees]. And you can forget about the z direction, this is a 2D problem.

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[quote]zImage wrote:
At 5 seconds or 1/4 the way around the circle, wouldn’t you be at <3,3,0> relative to O?

I don’t think you can use Cartesian velocity equations for this. The problem gives you angular velocity (1 rev/20 sec), you should just use that to determine position, for example:

at 5 seconds, (1 rev / 20 sec) = (X rev / 5 secs), now solve for X to get 0.25 revolutions, or 90 degrees.
at 7.5 seconds, (1 rev / 20 sec) = (X rev / 7.5 secs), now solve for X to get 0.375 revolutions, 135 degrees.
etc…

You only need arc length stuff for part d.

Good luck!
[/quote]

Those aren’t position vectors. Also, at t=7.5, the angle would be 67.5 degrees, not 135. The angle is wrt point O, not the center of the circle.

You don’t need the arc length stuff for part D either, a^2+b^2=c^2 that shit. The displacement would be [-3rt(2),-45deg]

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Thanks for help, I continued onto part b, could you tell me what I am doing wrong?

speed = 18degrees/second, distance = rt = (18deg/s)x(7.5sec) = 135degrees

-90 + 135 = 45 degrees. y = 3 sin 45, x = 3 cos 45. So if the circle was centered around the origin, the particle would be located at the point [1.5sqrt(2),1.5sqrt(2)]. Since the circle is shifted up 3 meters on the y axis, the particle is now located at [1.5sqrt(2),3 + 1.5sqrt(2)]

So using the distance formula, the distance from (0,0) to [1.5sqrt(2),3 + 1.5sqrt(2)] is sqrt[18 +9sqrt(2)]

But now I don’t know how to find the angle. I tried using the dot product with the x-axis somehow, but couldn’t get the right answer. This answer doesn’t even feel right, am I doing it wrong?

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[quote]grandin11 wrote:

[quote]zImage wrote:
At 5 seconds or 1/4 the way around the circle, wouldn’t you be at <3,3,0> relative to O?

I don’t think you can use Cartesian velocity equations for this. The problem gives you angular velocity (1 rev/20 sec), you should just use that to determine position, for example:

at 5 seconds, (1 rev / 20 sec) = (X rev / 5 secs), now solve for X to get 0.25 revolutions, or 90 degrees.
at 7.5 seconds, (1 rev / 20 sec) = (X rev / 7.5 secs), now solve for X to get 0.375 revolutions, 135 degrees.
etc…

You only need arc length stuff for part d.

Good luck!
[/quote]

Those aren’t position vectors. Also, at t=7.5, the angle would be 67.5 degrees, not 135. The angle is wrt point O, not the center of the circle.

You don’t need the arc length stuff for part D either, a^2+b^2=c^2 that shit. The displacement would be [-3rt(2),45deg][/quote]

I disagree, I got the magnitude-angle vector to be <3rt(2),45deg>, not <-3rt(2),45deg>. The negative would put you in the wrong quadrant.

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785×699 40.6 KB

I don’t think I was being very clear. Let me try to attach a calc

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[quote]jdinatale wrote:

[quote]grandin11 wrote:

[quote]zImage wrote:
At 5 seconds or 1/4 the way around the circle, wouldn’t you be at <3,3,0> relative to O?

I don’t think you can use Cartesian velocity equations for this. The problem gives you angular velocity (1 rev/20 sec), you should just use that to determine position, for example:

at 5 seconds, (1 rev / 20 sec) = (X rev / 5 secs), now solve for X to get 0.25 revolutions, or 90 degrees.
at 7.5 seconds, (1 rev / 20 sec) = (X rev / 7.5 secs), now solve for X to get 0.375 revolutions, 135 degrees.
etc…

You only need arc length stuff for part d.

Good luck!
[/quote]

Those aren’t position vectors. Also, at t=7.5, the angle would be 67.5 degrees, not 135. The angle is wrt point O, not the center of the circle.

You don’t need the arc length stuff for part D either, a^2+b^2=c^2 that shit. The displacement would be [-3rt(2),45deg][/quote]

I disagree, I got the magnitude-angle vector to be <3rt(2),45deg>, not <-3rt(2),45deg>. The negative would put you in the wrong quadrant.[/quote]

Well there’s two ways to go about doing it. Negative magnitude, positive direction or positive magnitude negative direction. I forgot to put a - in front of the 45, that way it would be correct. Now that I think about it, I would go with <3rt(2), 135deg>. Positive magnitude and direction (like what you have) would take you in the exact opposite direction from the center of the circle.

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[quote]grandin11 wrote:

[quote]jdinatale wrote:

[quote]grandin11 wrote:

[quote]zImage wrote:
At 5 seconds or 1/4 the way around the circle, wouldn’t you be at <3,3,0> relative to O?

I don’t think you can use Cartesian velocity equations for this. The problem gives you angular velocity (1 rev/20 sec), you should just use that to determine position, for example:

at 5 seconds, (1 rev / 20 sec) = (X rev / 5 secs), now solve for X to get 0.25 revolutions, or 90 degrees.
at 7.5 seconds, (1 rev / 20 sec) = (X rev / 7.5 secs), now solve for X to get 0.375 revolutions, 135 degrees.
etc…

You only need arc length stuff for part d.

Good luck!
[/quote]

Those aren’t position vectors. Also, at t=7.5, the angle would be 67.5 degrees, not 135. The angle is wrt point O, not the center of the circle.

You don’t need the arc length stuff for part D either, a^2+b^2=c^2 that shit. The displacement would be [-3rt(2),45deg][/quote]

I disagree, I got the magnitude-angle vector to be <3rt(2),45deg>, not <-3rt(2),45deg>. The negative would put you in the wrong quadrant.[/quote]

Well there’s two ways to go about doing it. Negative magnitude, positive direction or positive magnitude negative direction. I forgot to put a - in front of the 45, that way it would be correct. Now that I think about it, I would go with <3rt(2), 135deg>. Positive magnitude and direction (like what you have) would take you in the exact opposite direction from the center of the circle.[/quote]

I’m still disagreeing with you, if you went 45 degrees from the origin with a radius of 3rt(2), you will be exactly where you need to be. I don’t understand why you would want to use 135 degrees, because 135 degrees from the origin will place you in the 2nd quadrant, which is not what we want. Remember, you are using the angle from the ORIGIN, not from the center of the circle.

[quote]zImage wrote:
I don’t think I was being very clear. Let me try to attach a calc
[/quote]

Wow I did mine a MUCH harder way, and got the same answer…I wish i would have known your method! Thanks

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[quote]jdinatale wrote:

[quote]grandin11 wrote:

[quote]jdinatale wrote:

[quote]grandin11 wrote:

[quote]zImage wrote:
At 5 seconds or 1/4 the way around the circle, wouldn’t you be at <3,3,0> relative to O?

I don’t think you can use Cartesian velocity equations for this. The problem gives you angular velocity (1 rev/20 sec), you should just use that to determine position, for example:

at 5 seconds, (1 rev / 20 sec) = (X rev / 5 secs), now solve for X to get 0.25 revolutions, or 90 degrees.
at 7.5 seconds, (1 rev / 20 sec) = (X rev / 7.5 secs), now solve for X to get 0.375 revolutions, 135 degrees.
etc…

You only need arc length stuff for part d.

Good luck!
[/quote]

Those aren’t position vectors. Also, at t=7.5, the angle would be 67.5 degrees, not 135. The angle is wrt point O, not the center of the circle.

You don’t need the arc length stuff for part D either, a^2+b^2=c^2 that shit. The displacement would be [-3rt(2),45deg][/quote]

I disagree, I got the magnitude-angle vector to be <3rt(2),45deg>, not <-3rt(2),45deg>. The negative would put you in the wrong quadrant.[/quote]

Well there’s two ways to go about doing it. Negative magnitude, positive direction or positive magnitude negative direction. I forgot to put a - in front of the 45, that way it would be correct. Now that I think about it, I would go with <3rt(2), 135deg>. Positive magnitude and direction (like what you have) would take you in the exact opposite direction from the center of the circle.[/quote]

I’m still disagreeing with you, if you went 45 degrees from the origin with a radius of 3rt(2), you will be exactly where you need to be. I don’t understand why you would want to use 135 degrees, because 135 degrees from the origin will place you in the 2nd quadrant, which is not what we want. Remember, you are using the angle from the ORIGIN, not from the center of the circle.

[quote]zImage wrote:
I don’t think I was being very clear. Let me try to attach a calc
[/quote]

Wow I did mine a MUCH harder way, and got the same answer…I wish i would have known your method! Thanks
[/quote]

I’m talking about part d with that calculation…so it’s displacement in that interval. The origin has nothing to do with this part of the question, displacement is change in position. Meaning, it’s simply the the location of the final position wrt it’s initial position. I stand by what I said. Try drawing it out if you disagree.

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I agree grandin, my bad on that.

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Fuck it, beats studying for chem so here’s an image, hope it helps. The angle can’t actually read 45 degrees though, because that would be completely against the coordinate system you have set up.

My paint skills are severely lacking, as you can see. The red is the shit you don’t care about at all for this problem.

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I’m sorry Gradin11, I thought you were talking about a different part of the problem. For part D, I did this:

I don’t think I’m doing it right…I’m doing:

Displacement = net change in position = r(10)-r(5) = <0,6,0> - <3,3,0> = <-3,3,0>

I don’t know if that was the correct way of answering that question. Should I be getting a vector answer? Do I need to find the magnitude of that vector?

I would think that the “state the following vectors in magnitude-angle notation” only applied to part a, b, c?

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[quote]jdinatale wrote:
I’m sorry Gradin11, I thought you were talking about a different part of the problem. For part D, I did this:

I don’t think I’m doing it right…I’m doing:

Displacement = net change in position = r(10)-r(5) = <0,6,0> - <3,3,0> = <-3,3,0>

I don’t know if that was the correct way of answering that question. Should I be getting a vector answer? Do I need to find the magnitude of that vector?

I would think that the “state the following vectors in magnitude-angle notation” only applied to part a, b, c?[/quote]

Yep, that’s the right answer. I took it to mean that all answers were to be in magnitude-direction notation though, so maybe give that answer then just show calculations to put it into the right format. If you’re unclear and it’s easy enough, might as well give both answers, they won’t take marks off for that.

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For part D, in magnitude-angle displacement, I’m getting <3rt(2), 135 degrees>, I don’t see where you get -3rt(2), because if you go 135 degrees from the horizontal at t=5second, you need to go 3rt(2) in radius to get to the point at t=10 seconds.

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[quote]jdinatale wrote:
For part D, in magnitude-angle displacement, I’m getting <3rt(2), 135 degrees>, I don’t see where you get -3rt(2), because if you go 135 degrees from the horizontal at t=5second, you need to go 3rt(2) in radius to get to the point at t=10 seconds.[/quote]
I mentioned that earlier. Two ways here…negative mag, pos dir or pos mag, neg direction…I agree with your answer, and said that earlier. For the -3rt(2), the angle would have to be -45, or 315.

Hope that clears things up.