Confusing circular motion

[LondonLady]

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

\displaystyle{\frac{d\theta}{dt} = \omega}

Now assume that the circle has radius r and that

\displaystyle{\frac{d\theta}{dt} = 2t}

for t in seconds. Let \theta(t = 0) = 0

(therefore \theta = t^2)

a) Find \vec{r}(t)

b) Find \vec{v}(t). is \vec{v} \perp \vec{r}?

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

d) With respect to the circle's centre, sketch \vec{r},\vec{v} and \vec{a} 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 \vec{r} \perp \vec{v} and that \vec{a} \perp \vec{v}. I have also found that \vec{a} can be written as -\alpha \vec{r} (where \alpha is a constant. All this implies that the acceleration vector is pointed back into the centre of the circle as some negative multiple of \vec{r}.

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?

[arildno]

The position vector is perpendicular to the velocity vector, but the acceleration vector is NOT perpendicular to the velocity vector!

[LondonLady]

hmm... thankyou for your reply. I cant agree though. I got

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

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

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

Then if you find the dot product

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

Which implies they are perpendicular.... Is my logic wrong?

[arildno]

You have not differentiated \vec{v} correctly:
\frac{d\vec{v}}{dt}=(\frac{d}{dt}2tr)(-\sin(t^{2})\vec{i}+\cos(t^{2})\vec{j})+2tr\frac{d} {dt}(-\sin(t^{2})\vec{i}+\cos(t^{2})\vec{j})=
2r(-\sin(t^{2})\vec{i}+\cos(t^{2})\vec{j})+4t^{2}r(-\cos(t^{2})\vec{i}-\sin(t^{2})\vec{j})

[LondonLady]

Ahh! It didnt even occur to me that it might have changed into a product! Thankyou so much!