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Uniform circular motion - 2 dimensional - 2 forces

  1. Apr 12, 2012 #1
    An object moves in the xy-plane in a uniform circular motion. We know that there is two forces action upon the object. One we have that is F1, F2 is unknown. Both vectors

    speed : 4m/s
    radius : 4m
    mass : 1kg
    F1 : 10 i N (i as in x-direction vector)
    F2 : ? (bound to be in the xy-plane)


    How do I calculate F2, and I have to calculate it when the angle between the rope and the x-axis is 0, Pi/2 , Pi. The rope has no mass by the way. The motion is counter-clockwise, and the angle is zero when the ball is at the positive side of the x-axis (given that origin is in the middle of the motion).


    Thank you!

    Excuse my poor english
     
  2. jcsd
  3. Apr 12, 2012 #2
    Can you tell us what the acceleration of particle if it is in a Uniform circular motion?
     
  4. Apr 12, 2012 #3
    Hey finitefemmet,

    As the speed is given you can calculate the centripetal force required for the circular motion. Then find F2 such that F1 + F2 is equal to the required force and in the given direction.

    Try solving after this.
     
  5. Apr 12, 2012 #4
    Okey,

    so I have calculated that the net force for the circular motion is 4N.... -4(j) if we are at angle zero position.

    Now correct me if I am wrong on this:

    When the angle is 0, that is when the ball is on the right side of origin on the x-axis.
    The force is then pointing too origin and that is negative x-direction. F2 is then : -4(j) - 10 (i).

    Minus 10 too "cancel" out the y-axis force and -4(j) too get the right net force.

    And so on for the rest of the angles

    Right track?

    Thanks for the quick response!
     
  6. Apr 14, 2012 #5
    Is that correct?
     
  7. Apr 15, 2012 #6
    Well

    F = (m*v2) / r
    F = (1*4^2) / 4 = 4
    And the force is pointing towards the center.
     
  8. Apr 15, 2012 #7
    that is why it is -4(i) not (j)
     
  9. Apr 16, 2012 #8
    My mistake, you are absolutely right!

    But now that is solved, the rest of my thinking should be okey?
     
  10. Apr 16, 2012 #9
    yeah rest is right. The only mistake was this one
     
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