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Problem with motion in a circular path

  • Thread starter Heather
  • Start date
  • #1
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Homework Statement


Calculate the average vector velocity between 0 and pi/4 sec.

Homework Equations


x=rcos2t
y=rsin2t
Vx=2rsin(2t)
Vy=2rcos2t
Ax=4rcos(2t)
Ay=-4cos(2t)
Circular path is x^2+y^2+r^2

The Attempt at a Solution


I'm not sure if I am missing something simple or not, What I need to know is what is r and so I initially set the square of the x and the square of the y equations equal to the square r. Pythagorean style. when I solved for r I felt just a little dumb because I got 1=1 which well duh I shoul've saw that coming. Now I'm trying to guess whether I even need r to do an average, or if there is another way to solve for r that I am missing. I also know that x=r when t=0 but when I sub r for x in the position equation I just get 1.
 

Answers and Replies

  • #2
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What are the x and y coordinates of the particle at time t = 0?
What are the x and y coordinates of the particle at time t = π/4?
What is the vector displacement of the particle (in terms of the unit vectors in the x and y directions) between time t = 0 and time t = π/4?
What is the definition of average (vector) velocity?

Chet
 
  • #3
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At t=0 y=0 and x=r
At t=π/4 I an uncertain because I haven't been able to figure at what r is.
I guess my real problem isn't finding the average but figuring out what r is.
 
  • #4
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At t=0 y=0 and x=r
At t=π/4 I an uncertain because I haven't been able to figure at what r is.
I guess my real problem isn't finding the average but figuring out what r is.
r is the radius of the circle, which is treated as an algebraic parameter.

chet
 
  • #5
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ahh duh, ok so then it should look something like this, taking the root of the sum of the x and y coordinates for t=π/4 and the root of the sum of the x and y coordinates at t=0 subtract the final position from the initial, and divide by two. Giving (2-root2)/2.
 
  • #6
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That's not correct. But, it's hard to tell what you did. Please, just answer, in order, the 4 questions I asked in post #2.

Chet
 
  • #7
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Is 0.7458 correct?

#1 at t=0, x=r and y=0
#2 at t=pi/4, x=rcos(2pi/4) and y=rsin(2pi/4)
#3 Displacement at x=rcos(2pi/4)-x(or r) and y=rsin (2pi/4) - 0
#4 The average vector velocity is the (final position - initial position) dived by the (final time - initial time).
 
  • #8
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4,096
[
Is 0.7458 correct?

#1 at t=0, x=r and y=0
#2 at t=pi/4, x=rcos(2pi/4) and y=rsin(2pi/4)
What is cos(2π/4) equal to?
What is sin(2π/4) equal to?
#3 Displacement at x=rcos(2pi/4)-x(or r) and y=rsin (2pi/4) - 0
Please express this in vector form using unit vectors or with an ordered pair of components.
#4 The average vector velocity is the (final position - initial position) dived by the (final time - initial time).
please express this in vector form using unit vectors or with an ordered pair of components.
What is the magnitude of this average velocity vector?

Chet
 
  • #9
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#3 Position= r(0)i+r(1)j for final, and Position=r(1)i+r(0)j for initial

#4 ||v||=√((-1r/(π/4))2+(1r/(π/4))2) --> (4 √(2) √(r2))/pi
 
  • #10
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4,096
#3 Position= r(0)i+r(1)j for final, and Position=r(1)i+r(0)j for initial
This should be written as
Initial position = r i
Final position = r j

Displacement vector = - r i + r j
#4 ||v||=√((-1r/(π/4))2+(1r/(π/4))2) --> (4 √(2) √(r2))/pi
What is ##\sqrt{r^2}## equal to?

They did not ask for the magnitude of the average velocity vector. They asked for the average velocity vector itself. This is equal to the displacement vector divided by the time interval pi/4. What do you get for that?

Chet
 
  • #11
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The average velocity vector would be v=-(4r/π)i+(4r/π)j
 
  • #12
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4,096
The average velocity vector would be v=-(4r/π)i+(4r/π)j
OK. We're done here. Nice job.

Chet
 
  • #13
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yes! thank you so much for helping me through that problem, and for being so patient.
 

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