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Circular motion-really need some help

  1. May 2, 2006 #1
    hi

    I'm stuck so I'd like to ask a couple of questions about circular motion.

    1*I know that for certain radius R and certain centripetal acceleration vector ( centr. acc. vector ), magnitude of velocity has to be just right for object ( either point mass or rigid body ) to go in circles. Because then acc. vector is perpendicular to velocity vector and thus it only changes its direction, but not its magnitude. But why does magnitude of velocity have to be just right for acc. vector to be perpendicular to velocity?

    Why if object's velocity is too slow are centr. force and thus centr. acc not perpendicular to velocity?

    Does acc.vector ( at the moment magnitude of velocity gets too slow for object to circle ) pull object in such position that cent. force is no longer perpendicular to object, and because of that object goes in some other, non circular path?






    2*In first question I stated that for object to go in circles ( at known R and centr. acc. vector ), it must have certain speed. But that obviously is not true when acc. vector also has tangential component aka angular acc.

    So how does object manage to go in circles even though the magnitude of its velocity is either increasing or decreasing?


    Is it because component ( of acc vector ) perpendicular to velocity also keeps changing ? If magnitude of velocity keeps getting bigger, then in order for an object to keep going in circles, acc vector directed towards the center of circle must also keep getting bigger (i.e. centripetal force must keep getting bigger as long as magnitude of velocity is getting bigger ). And reason for that is because greater the magnitude of velocity vector, the greater must its change in direction be?


    thank you
     
  2. jcsd
  3. May 2, 2006 #2

    Andrew Mason

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    Velocity has to be 'just right' only if the force cannot adjust. This is the case if the central force is gravity, for example. If the central force is provided a string or a roller coaster track which has a fixed radius, the velocity can be anything and it is the central force (acceleration) that 'adjusts'. Actually the force is simply determined by the rate of change of velocity. The rate of change of velocity is determined by the path (which is determined by the string or track) and its speed.

    I would suggest that you work through your text book and work it out from first principles: f=ma.

    AM
     
  4. May 2, 2006 #3

    daniel_i_l

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    The faster you spin the mass, the more force you need to keep it rotating in a circle. (cause the centripital force gets bigger)
     
  5. May 3, 2006 #4

    I know how we derived F=m*r*w^2 from F=m*a and I also get the basic idea behind the formulas.If you read my questions again perhaps you will notice that my questions are a bit "deeper".
     
  6. May 3, 2006 #5

    nrqed

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    Consider a force of constant magnitude. Then you can visualize why the speed has to be just "right" to keep the object moving along a circle. In a certain time interval [itex] \delta t [/itex] the force will pull the object toward the center. But the object has an initial velocity so it will also move in the direction of its initial velocity. Imagine that we use an x-y coordinate sytem with the origin located at the center of the Earth and imagine that we start looking at the object when it is on the y axis (so just "above" the Earth). A little time interval later, the object will have "fallen" a bit along the y axis but it will also have moved a bit on the x axis. If the speed is "just right", the object will still be at the same distance as it initially was from the center of the Earth, hence it follows a circle. (Of course, everything should really be done in the limit of infinitesimal delta t).


    If the object moves above the "right speed", the it will have moved more along the x axis then in the previous example and its distance from the Earth will be greater than itstarted with so it it is not following a circle.

    It is hard to explain clearly without a blackboard but do you see what I mean?
     
  7. May 3, 2006 #6

    Andrew Mason

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    Nothing in this is deeper than f=ma. Everything follows from f=ma. That is my point.

    AM
     
  8. May 3, 2006 #7

    Curious3141

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    A lot of the confusion surrounding circular motion (IMHO) comes from the way in which the introductory proof of formuale like [tex]F = mr\omega^2 = \frac{mv^2}{r}[/tex] is presented.

    Personally, the only proof of the acceleration experienced by a body in circular motion that I approve of is the rigorous one using polar or complex vector representation that in a few simple mathematical manipulations tells us *all* that we need to know about circular motion.

    [tex]z = re^{i\theta}[/tex] where z represents the complex position vector of a particle. Real parts represent radial components, imaginary parts represent tangential components.

    [tex]v = \frac{dz}{dt} = (i)r\dot{\theta}e^{i\theta} + \dot{r}e^{i\theta} = \dot{r}(\hat{z}) + i(r\omega)(\hat{z})[/tex]

    and there we have the formulae for the radial component of the velocity and the tangential component, respectively.

    [tex]a = \frac{dv}{dt} = -r{\dot{\theta}}^2e^{i\theta} + ir\ddot{\theta}e^{i\theta} + i\dot{r}\dot{\theta}e^{i\theta} + \ddot{r}e^{i\theta} + i\dot{r}\dot{\theta}e^{i\theta} = (-r\omega^2 + \ddot{r})(\hat{z}) + i(2\dot{r}{\omega} + r\alpha)(\hat{z})[/tex]

    and there we have represented the radial components for centripetal acceleration and direct radial acceleration, and the tangential components for the Coriolis-type acceleration and tangential acceleration due to angular acceleration.

    I consider that to be a direct and rigorous proof instead of the wishy-washy handwaving proofs in Basic Physics texts involving triangles and small angle approximations. And it's efficient, it encapsulates all that you need to prove in a few short lines.
     
    Last edited: May 3, 2006
  9. May 3, 2006 #8
    Below I am assuming object orbiting around the earth has circular orbit!


    Biggest problem I have with gravity as being centripetal force for orbiting object is that when talking about centripetal force I tend to think of it as force that simply changes direction of velocity vector.But reason for object orbiting the earth is because the object falls towards the earth at the same rate that the earth curves.So from this description one doesn't get the feeling that gravity force changes direction of velocity, but instead that object just happens to have enough speed for it to fall at same rate as earth curves.

    I know gravitational force causes falling, but I just can't seem to interpret it as centripetal force even if effects(object going in circles) are the same in both cases
     
    Last edited: May 3, 2006
  10. May 3, 2006 #9

    Andrew Mason

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    The force is always toward the centre of the earth. The velocity is always tangential to the orbit. If the orbit is a circle, the force is always perpendicular to the velocity. Draw a vector whose length is proportional to speed and direction is tangential to the orbit. Draw a vector that represents its speed a small time later. Draw a vector from the end of the first one to the end of the second one. That little vector is the change in velocity. Where does it point? How does it compare to the direction of the force of gravity? That change in velocity is the centripetal acceleration.

    AM
     
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