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Task-Uniform Circular Motion

  1. Oct 28, 2011 #1
    I have a question for you how to slove a task in physic
    So this task is about Uniform Circular Motion
    r=4 m
    T=6 s
    m=1.5 kg angle=50 degree

    I have to find as a distance
    x=? y=?

    velocity acceleration force
    V=? a=? F
    Vx=? ax=? Fx
    Vy=? ay=? Fy
     
  2. jcsd
  3. Oct 28, 2011 #2
  4. Oct 29, 2011 #3
    A little respect never hurt anybody
     
  5. Oct 29, 2011 #4

    Simon Bridge

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    No need for that language: it's only been an hour and a half.
    I'm not going to give you the answers - just point you in a useful direction.

    You clearly do not know what those equations mean.

    Lets go back to the first post and go through the things you need to know:

    r = radius: the distance of the object from the center of the circle
    T = period: this is the time it takes the object to go right around the circle.
    m = mass of the object

    important concepts:
    angular velocity: [tex]\omega = \frac{d\theta}{dt}[/tex] ... for uniform circular motion, this is a constant.

    That the object goes all around the circle in time T means that it goes through an angle of 2\pi radiens in T seconds so that [tex]\omega = \frac{2\pi}{T}[/tex] radiens per second. At any time t, the angle is [tex]\theta(t)=\omega t[/tex]

    Starting to get it?

    If you put cartesian axis on your circular path, you can find the x-y coordinates by trigonometry.

    You get the x and y velocities by differentiating the equations for x and y wrt time, and similarly for x and y accelerations.

    The linear speed that the object goes around the circle though is a constant - it is just distance over time ... and you know that it goes the whole circumference of the circle in one time period.

    What's important though is the centripetal acceleration which is [tex]a_c = \frac{v^2}{r}[/tex] The centripetal force is given by this times the mass. This is the force holding the object on to the circle.

    You should be able to get the rest from there.
     
  6. Oct 30, 2011 #5

    Simon Bridge

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    So anyway ... you feel better about circular motion now?
     
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