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Homework Help: Confusing circular motion

  1. Nov 14, 2004 #1
    Im a bit confused about a question on circular motion that i'm answering. Ill state the entire question and then say what im confused about.

    In class we discussed circular motion for the case

    [tex]\displaystyle{\frac{d\theta}{dt} = \omega}[/tex]

    Now assume that the circle has radius [tex]r[/tex] and that

    [tex]\displaystyle{\frac{d\theta}{dt} = 2t}[/tex]

    for [tex]t[/tex] in seconds. Let [tex]\theta(t = 0) = 0[/tex]

    (therefore [tex]\theta = t^2[/tex])

    a) Find [tex]\vec{r}(t)[/tex]

    b) Find [tex]\vec{v}(t)[/tex]. is [tex]\vec{v} \perp \vec{r}[/tex]?

    c) Find [tex]\vec{a}(t)[/tex]. Express [tex]\vec{a}[/tex] in terms of [tex]\vec{r}[/tex] and [tex]\vec{v}[/tex]. Is [tex]\vec{a} \perp \vec{v}[/tex]?

    d) With respect to the circle's centre, sketch [tex]\vec{r},\vec{v}[/tex] and [tex]\vec{a}[/tex] for counter clockwise rotation.

    Ok. I have found all the vectors in i-j form. My question is about the perpendicularity questions. Mathematically I have found that [tex]\vec{r} \perp \vec{v}[/tex] and that [tex]\vec{a} \perp \vec{v}[/tex]. I have also found that [tex]\vec{a}[/tex] can be written as [tex]-\alpha \vec{r}[/tex] (where [tex]\alpha[/tex] is a constant. All this implies that the acceleration vector is pointed back into the centre of the circle as some negative multiple of [tex]\vec{r}[/tex].

    If this is so then how is the particle speeding up?? (the rate of change of theta is time dependent)

    I would have thought that the acceleration vector would have been at some angle to the position vector. But then it wouldnt move in a circle.... Im confused....

    Also, the second part of (c) im finding hard. Anyone any ideas?
  2. jcsd
  3. Nov 14, 2004 #2


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    The position vector is perpendicular to the velocity vector, but the acceleration vector is NOT perpendicular to the velocity vector!
  4. Nov 14, 2004 #3
    hmm... thankyou for your reply. I cant agree though. I got

    [tex]\vec{r}(t) = r\cos (t^2)\vec{i} + r\sin (t^2)\vec{j}[/tex]

    [tex]\vec{v}(t) = -2tr\sin (t^2)\vec{i} + 2tr\cos (t^2)\vec{j}[/tex]

    [tex]\vec{a}(t) = -4t^2r\cos(t^2)\vec{i} - 4t^2r\sin(t^2)\vec{j}[/tex]

    Then if you find the dot product

    [tex]\vec{a}.\vec{v} = 8t^3r^2\sin(t^2)\cos(t^2) - 8t^3r^2\sin(t^2)\cos(t^2) = 0[/tex]

    Which implies they are perpendicular.... Is my logic wrong?
  5. Nov 14, 2004 #4


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    You have not differentiated [tex]\vec{v}[/tex] correctly:
  6. Nov 15, 2004 #5
    Ahh! It didnt even occur to me that it might have changed into a product! Thankyou so much!
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