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Motion of a particle -xy plane

  1. Mar 8, 2007 #1
    1. The problem statement, all variables and given/known data
    The motion of a particle moving in a circle in the x-y plane is described by the equations: r(t)=9.71, theta(t)=8.74t. Where theta is the polar angle measured counter clockwise from the + x-axis in radians, and r is the distance from the origin in m.
    A. Calculate the y-coordinate of the particle at the time 1.60s.
    B. Calculate the y-component of the velocity at the time 3.40s.
    C. Calculate the magnitude of the acceleration of the particle at the time 1s.
    D. By how much does the speed of the particle change from t=10s to t=67s?
    E. Calculate the x-component of the acceleration at the time 2.20s.

    2. Relevant equations


    3. The attempt at a solution
    I usually try to make several attempts before I ask for help, but I don't even know where to begin! How do I go about this 5 part problem? Please help me :confused:
  2. jcsd
  3. Mar 8, 2007 #2


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    Do you know expressions relating polar coordinates to cartesian coordinates? If not, try drawing a diagram!
  4. Mar 8, 2007 #3
    I just know that x= r cos theta and y= r sin theta
    So...to find part A I do y= r(t)sin (theta(t))?
    Unfortunately, I don't see how I can use that for parts B through E.
  5. Mar 8, 2007 #4


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    Well, how are position and velocity related?
  6. Mar 8, 2007 #5
    I know that velocity= distance/time...
    I just tried part A and got that incorrect. I did y=r(t)*sin(theta(t)). Why is that incorrect?
  7. Mar 8, 2007 #6


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    Until you present your work, I'll be unable to conclude why it's incorrect.
  8. Mar 8, 2007 #7
    when I tried part A I put in y= 9.71*sin (8.74*1.60)
  9. Mar 8, 2007 #8


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    Is your calculator set to radians?
  10. Mar 8, 2007 #9
    It wasn't. I just switched it to radians.
  11. Mar 8, 2007 #10
    Now I got the correct answer for part A. Thank you for pointing that out.

    I know you wrote about the relationship between velocity and position but I cannot figure out the relationship in this coordinate system... I understand for poisition in the y component, it will be r*sin(theta(t)), but I'm lost after that.

    I understand for part C that I need to find the x and y components of acceleration in order to find the magnitude of acceleration...but how do I figure out the equation for acceleration for this coordinate system?
  12. Mar 8, 2007 #11


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    Again, look at post #4, look at your answer at post #5, and try to give a precise relation.
  13. Mar 8, 2007 #12
    Radou, I'm sorry if I am slow at figuring out how to do this problem!

    I looked at post 4 and 5 again... since velocity=distance/time, then I can find velocity of the y component using Vy = r*sin(theta(t)/ t

    and since acceleration is equal to the change in velocity divided by the change in time, then Ax= (r*cos(theta(t))/t)/t
    Ay= (r*sin(theta (t))/t)/t
    and proceed from there.
  14. Mar 8, 2007 #13


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    No need to apologize.

    I suggest you look up the term 'instantaneous velocity'.
  15. Mar 9, 2007 #14
    Before I viewed your response I attempted v= r*sintheta(t)/t and I got an incorrect answer.

    I found out that the angle theta increases linearly with time so the speed v is constant and is given by r*d(theta)/dt and that r is constant in problem.
    Which I think deals with what you said...instantaneous velocity.

    So since this is dealing with derivative, then theta(t) becomes just theta and t goes away? Then Vy=r*theta. Or is it Vy=r*theta(t)/t?
  16. Mar 9, 2007 #15
    I attempted Vy=(9.71m)*(8.74*3.40s)= 84.86 m/s. But that is incorrect.
    So after looking at your last post radou, I've been looking at the concept of instantaneous velocity. I understand its the first derivative of a position equation. Which is why at one point I thought v=r*theta which turns out to be 84.86 m/s as well.

    Then I realized that it is still a y component, so Vy=r*sintheta* d(theta)/dt
    I then did Vy=9.71*sin(8.74*3.40s)*(8.74*3.40)/3.40= -84.15m/s and that was incorrect. Can someone please tell me why that is incorrect?
  17. Mar 10, 2007 #16


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    Hint: chain rule for differetiation.
  18. Mar 10, 2007 #17

    Gib Z

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    When using calculus velocity is much better defined as [tex]V=\frac{ds}{dt}[/tex] rather than distance/time.
  19. Mar 10, 2007 #18
    radou, if I differentiate the equation theta(t), then it would be just theta, right? So then Vy=r*sin(theta(t)*theta. But when I put (9.71)*sin(8.74*3.40)*8.74=Vy, Vy still ends up equalling -84.15 m/s.

    Is my differentiation wrong?

    Gib Z, I will try what you suggested, but what happens to r? Since s=r*theta,

    would d(theta)/dt= 1/r *ds/dt?
  20. Mar 10, 2007 #19
    Ok, so I read more on differentiation and instantaneous velocity, etc.
    And I was right that d(theta)/dt ends up being theta.
    I went on to solve part c and part e.
    Yet I cannot solve part b. Once again, I put in r*sintheta(t)* theta=-9.71*sin(8.74*3.40)*8.74= 84.15 m/s and that of course was incorrect.

    I now know I am on the right track since I solved part e doing Ax= -theta^2*r*costheta(t).

    What am I doing wrong?:bugeye:
  21. Mar 10, 2007 #20


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    It seems your differentiation is wrong.
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