How do vectors relate to perpendicularity in 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 I am 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 wouldn't move in a circle... I am confused...


Also, the second part of (c) I am 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 can't 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!