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Urgent: Uniform Circular Motion Questions

  1. Mar 19, 2012 #1
    I need these Questions answers in 2 hours! :p Any help would be greatly appreciated.

    1. The problem statement, all variables and given/known data
    Question 3

    Mathematically derive a relationship between the speed v, of a body moving in a circle of radius r, and the frequency f, of the revolutions.

    Question 4
    Use your results for Q3 to next deduce a relationship between Fc and the frequency f. Thus, predict the shape of a graph of Fc against f and the shape of a graph of Fc against f2. State which variables must be held constant in this analysis.

    2. Relevant equations

    3. The attempt at a solution
    Question 3:

    I tried this:



    But I believe that answer is wrong. :/

    Question 4:
    Without knowing the above I can't work this out. I actually don't know how to work this out though. Any hints?

  2. jcsd
  3. Mar 19, 2012 #2
    Assuming the motion is uniform circular, use f = 1/T, v=r*angular velocity of the particle, and angular velocity of the particle = 2∏/T = 2∏f. That should solve Q3. The thing you have done incorrectly is that in circular motion, v or the linear speed represents the magnitude of the velocity of the revolving particle at an instant, with respect to the centre. In uniform circular motion, this speed is constant at all times, since angular velocity of revolution is constant at all times....

    Now Fc or the centripetal force = mv2/r at all instants during the motion of the particle. Now use the relations stated above to find the relationship between Fc, m (mass of he revolving particle), r and f. This ought to give you the required graph too. This solves Q4
  4. Mar 19, 2012 #3
    Thanks for your help. I still don't understand how you worked out Q3 though.
    I get f = 1 /2 pi r / v, :/

    Wikipedia says ω = 2πf is the angular frequency, is that what I want?

    The way I tried to do it was: T = 2pi r / v, and f = 1/T,

    So just replace T with 2pir /v, to get my formula which has f, r and v which are all needed. ω = 2πf doesn't satisfy the question.
  5. Mar 19, 2012 #4


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    That is correct, but it is a bit overcomplicated, as it is a fraction with a fraction in the denominator. How do you get the reciprocal of a fraction?

  6. Mar 19, 2012 #5
    The mistake you are making in Q3 is that you are writing the linear speed as 2∏r/T. This may sound like distance/time taken, but in fact, distance/time taken is termed as the average speed of the object for that time. Here, we are talking about the linear speed in the circular motion, that is the magnitude of the velocity of the revolving particle at any instant of time with respect to the centre of the circular path. This speed is not the 'average speed' you are using in your equation. Now this linear speed v = rω (ω=angular velocity of revolution = 2πf, as you said), at all instants in the uniform circular motion. This is what I meant.
  7. Mar 19, 2012 #6


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  8. Mar 19, 2012 #7
    Hey ehild,
    What I mean to say is that here, in case of uniform circular motion, we specifically say that the speed of the particle is constant with time. Here, the 'speed' we mean is the magnitude of velocity of the rotating particle wrt the centre at all times, and hence it is constant, as revolution angular velocity is constant.
  9. Mar 19, 2012 #8


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    Yes, and it is v=(circumference of the circle)/(time period) v=(2πr)/T as Mohammad said, and it is also v=(2πf)r=ωr.

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