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A bit of help please

  1. Sep 15, 2005 #1

    Boy can physics be confusing :(

    change of velocity - delta (V)
    Short time period - delta (T)
    Change of angle - delta (alpha)
    Velocity - V
    Centripetal acceleration - a

    This I understand

    V = radius * delta (angle) / delta (time)

    In circular motion speed velocity only changes direction but not size , but
    still I don't quite understand how

    delta (V) = V * delta (alpha)

    represents (directional) change of velocity , and thus also don't understand

    a = delta (V) / delta (T)

    thank you
  2. jcsd
  3. Sep 15, 2005 #2


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    From the second attachment you can see that
    [tex]\Delta\theta=\frac{\Delta v}{v}[/tex]
    is just the magnitude of the velocity. I think you can do the rest yourself?

    Attached Files:

  4. Sep 15, 2005 #3
    I see it now .
    I know that delta(v) is vector which tells us how coordinates of velocity have changed ( since magnitude stays the same ),but how do you show that in formula , meaning if you don't want to draw a coordinate system with velocity vectors in it ?
  5. Sep 16, 2005 #4


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    When dealing with equations one need to distinguish carefully between scalar and vector quantities. Handbooks use different notations to indicate scalar and vector quantities in equations like for instance
    [tex]\Delta v[/tex]
    would refer to the magnitude of the change in velocity, while
    [tex]\Delta \overrightarrow{v}[/tex]
    would refer to the vector of the change in the velocity of an object. The notation of your handbook might be different. Is this what was bothering you?
  6. Sep 16, 2005 #5
    I think so,since I'm assuming delta(v) in this context refers to vector and not magnitude . I mean magnitude of velocity stays the same so delta(v) has to represent a vector

    And I'm sorry I didn't answer your post sooner,I was shure there was no reply !
  7. Sep 17, 2005 #6


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    It's ok nobody does physics all the time, at least not I. If your book is dealing with vector equations then the next step that you mention above
    [tex]\overrightarrow{a}=\frac{\Delta\overrightarrow{v}}{\Delta t}[/tex]
    indicates that the vector acceleration is in the same direction as
    [tex]\Delta \overrightarrow{v}[/tex]
    which points to the centre of the rotation. We are therefore deriving the vector equation for the centripetal acceleration of the rotating object.
  8. Sep 17, 2005 #7
    Perhaps the following question would deserve a thread of its own

    I know that delta(v) tells by how much velocity vector has moved from previous position . But in the end it just tells you that velocity vector now has coordinates 4 units ( for example ) away .

    So how can you take that information and use it to find acceleration is beyond me . You're basicly saying the in specific time velocity vector changes position by 4 units ( be it centimeters or whatever ) ?!

    And how do we know that delta(v) is directed towards the center of a circle ( proof I mean ) ?

    thank you for putting up with me
  9. Sep 17, 2005 #8


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    Draw a diagram of these velocity vectors,
    one at time t and one at time t + Delta(t).
    (By the way, Delta(t) is the change of time)
    In circular motion, each velocity vector is
    perp to their radius vector. Subtract the vectors:
    Delta(v) = v(t+Dt) - v(t) , is perp to v.
  10. Sep 17, 2005 #9
    Remember this formula for the arc length of a circle?


    Take the derivative of each side with respect to t.
  11. Sep 18, 2005 #10


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    Vectors are somewhat similar to ordinary numbers in the sense that if you add/subtract vectors you get a vector which have the same meaning as the original vector. In our case
    [tex]\Delta \overrightarrow{v}[/tex]
    is also a velocity vector - it tells us by how much (size and direction) the velocity of the rotating object have changed during the time
    [tex]\Delta t[/tex]
    that is
    [tex]\Delta \overrightarrow{v}[/tex]
    is not a displacement vector! About the proof of the direction - it will require some quite advanced maths to do that, but intuitively you can see it in the second attachment that if the time interval becomes very short that our
    [tex]\Delta \overrightarrow{v}[/tex]
    vector will become more and more perpendicular to the velocity vector, which you can see from the first attachment points along the radius of the circle, which is the direction of the acceleration of the rotating object.
  12. Sep 21, 2005 #11
    Sorry but I've been away for the past few days

    So basicly in this particular case where magnitude stays the same delta(v) tells you by how many units velocity has changed ( and thus for example delta(v) would only be saying that velocity changed its direction by for units ) ?
  13. Sep 21, 2005 #12
    I think you are still trying to understand [tex]a=\frac{\Delta v}{\Delta t}[/tex] ?

    This is just the definition of acceleration, which is the mathematical way of saying that acceleration is how velocity changes over time.

    In the present case, your [tex]\Delta v[/tex] approximates the acceleration vector. As these elements shrink, it gets closer and closer to a vector perpendicular to v, which will also have a different magnitude.

    a (which [tex] \Delta v [/tex] approaches as its limit) will have both a different direction (perpendicular to v) and a different magnitude from v.
    Last edited: Sep 21, 2005
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