1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?