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Find the velocity and acceleration vectors for uniform circular motion

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration.
    Suppose that a particle's position is given by the following expression:
    %7D%28t%29%3D+R%5B%5Ccos%28%5Comega+t%29%5Chat%7Bi%7D+%2B+%5Csin%28%5Comega+t%29%5Chat%7Bj%7D%5D.gif
    2C%5C+%5Cqquad+%3D+R%5Ccos%28%5Comega+t%29%5Chat%7Bi%7D+%2B+R%5Csin%28%5Comega+t%29%5Chat%7Bj%7D.gif

    A.) Choose the answer that best completes the following sentence:
    The particle's motion at t=0 can be described by ____________.

    B.) When does the particle first cross the negative x axis?
    Express your answer in terms of some or all of the variables omega, R, and pi.

    C.) Find the particle's velocity as a function of time.
    Express your answer using unit vectors (e.g., A i_unit+ B j_unit, where A and B are functions of omega, R, t, and pi).

    D.) Find the speed of the particle at time t.
    Express your answer in terms of some or all of the variables omega, R, and pi.

    E.) Now find the acceleration of the particle.
    Express your answer using unit vectors (e.g., A i_unit+ B j_unit, where A and B are functions of omega, R, t, and pi).

    F.) Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r_vec(t).
    Express your answer in terms of some or all of the variables r_vec(t) and omega.

    H.) Finally, express the magnitude of the particle's acceleration in terms of R and v using the expression you obtained for the speed of the particle.
    Express your answer in terms of one or both of the variables R and v.

    3. The attempt at a solution
    I did parts A-D. But am stuck starting with Part E, and haven't really tried the rest
    below is the particle's velocity as a function of time. I know to get acceleration you're supposed to take the derivative of that, but I seem to be getting it wrong. If anyone could help me out I would appreciate it.

    t%29%5Chat%7Bi%7D%2B%7B%5Comega%7DR%7B%5Ccos%7D%5Cleft%28%7B%5Comega%7Dt%5Cright%29%5Chat%7Bj%7D.gif
     
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 24, 2010 #2

    rl.bhat

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    Homework Helper

    The question is not complete. Post the complete question.
     
  4. Jan 24, 2010 #3
    Alright sorry, whole question is up.
     
  5. Jan 24, 2010 #4
    Your method sounds good. What did you get for the second derivative of r(t)?
     
  6. Jan 24, 2010 #5
    derivative of r(t) is: t%29%5Chat%7Bi%7D%2B%7B%5Comega%7DR%7B%5Ccos%7D%5Cleft%28%7B%5Comega%7Dt%5Cright%29%5Chat%7Bj%7D.gif

    My main issue is I can't find the derivative of a(t) from the v(t) eq. or at least, I seem to be doing it wrong.
     
  7. Jan 24, 2010 #6
    You found the derivative of r(t):

    [tex]v(t) = \frac{dr}{dt} = -{\omega}R{\sin}\left({\omega}t\right)\hat{i}+{\omega}R{\cos}\left({\omega}t\right)\hat{j}[/tex]

    So find a(t)=dv/dt in just the same way.
     
  8. Jan 24, 2010 #7
    would the derivative be:
    ght%29%5Chat%7Bi%7D-R%7B%5Comega%7D%7B%5Csin%7D%5Cleft%28%7B%5Comega%7Dt%5Cright%29%5Chat%7Bj%7D.gif
     
  9. Jan 25, 2010 #8
    Almost. You forgot to bring out the omega. It might help to ignore the scalars (constants R and omega on the OUTSIDE of your functions) when taking the derivative. Try just

    [tex]\frac{d}{dt}( -{\sin}\left({\omega}t\right)\hat{i}+{\cos}\left({\omega}t\right)\hat{j})[/tex]

    then put the factors back in! (Hint: you should get omega squared!)
     
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