How do I find the centripetal acceleration when vectors are given?

Question

The problem is as follows:

A lactose solution is put in a test tube inside a centrifuge as shown in the graph from below where a small dot represents the tube with the sugar (view from the top). Assuming that the sugar sample along the test tube behaves as a particle, spins following a rotation with constant angular acceleration. The radius of the apparatus is $32\pi\,m$ and its angular acceleration is $\frac{\pi}{6}\,\hat{k}\,\frac{rad}{s^{2}}$. When $t=0\,s$ the tube is at point labeled $A$ and it has an angular speed of $\frac{\pi}{12}\,\hat{k}\,\frac{rad}{s}$. Find the centripetal acceleration when $t=4\,s$

The alternatives given on my book are:

$\begin{array}{ll} 1.&9\pi\left(-\hat{i}+\sqrt{3}\hat{j}\right)\\ 2.&4\pi\left(-\hat{i}+3\hat{j}\right)\\ 3.&9\pi\left(\hat{i}-\sqrt{3}\hat{j}\right)\\ 4.&4\pi\left(-\hat{i}-3\hat{j}\right)\\ 5.&3\pi\left(\hat{i}-\sqrt{3}\hat{j}\right)\\ \end{array}$

For this particular problem the only relevant equation for position which I can recall for rotation with constant angular acceleration is:

$\theta (t)=\theta_{o}+\omega_{0} t + \frac{1}{2}\alpha t^2$

But from then is where I'm stuck. I don't know how to relate this equation with the values that are given other than:

When $t=0$, then $\omega_{0}=\frac{\pi}{12}$

$\theta (t)=\theta_{o}+ \frac{\pi}{12} t + \frac{1}{2}\left(\frac{\pi}{6}\right) t^2$

$\theta (t)=\theta_{o}+ \frac{\pi}{12} t + \frac{\pi}{12}t^2$

$\theta(0)=\theta_{o}+ \frac{\pi}{12} (0) + \frac{\pi}{12}(0)^2$

$\theta(0)=\theta_{0}$

Edit:

I attempted to go further and I noticed the following:

Since what it is known is the angular acceleration, I can find the final angular speed and using this value I can obtain the centripetal acceleration which is what they are asking.

$\omega_{f} = \omega_{0} + \alpha t$

$\omega_{f} = \frac{\pi}{12} + \left(\frac{\pi}{6}\right)(4)$

$\omega_{f} = \frac{9\pi}{12}= \frac{3\pi}{4}$

From there:

$a_{c}=\left(\frac{3\pi}{4}\right)^2\left(\frac{32}{\pi}\right)$

$a_{c}=18\pi$

The position must be obtained from the position equation in rotational motion. In this case it has a constant acceleration (angular)

Now for this part I'm assuming when the object is at:

At $A$ then $\theta(0)=0$

$\theta(t)=\theta_{0}+\omega_{0} t + \frac{1}{2}\alpha t^2$

$0=\theta(0)=\theta_{0}+\omega_{0}(0) + \frac{1}{2}\alpha (0)^2$

$\theta_{0}=0$

Then this is simplified to:

$\theta(t)=\omega_{0} t + \frac{1}{2}\alpha t^2$

The initial speed is given:

$\theta(t)=\frac{\pi}{12} t + \frac{1}{2}\frac{\pi}{6} t^2$

$\theta(t)=\frac{\pi}{12} t + \frac{\pi}{12} t^2$

Then by pluggin in $t=4$

there should be:

$\theta(t)=\frac{\pi}{12} (4) + \frac{\pi}{12} (4)^2=\frac{5\pi}{3}$

Which gives:

$\theta=\frac{5\pi}{3}$

which is in the fourth quadrant.

Since what is being asked is the position all that would be left to do is to take the sine and cosines functions to obtain the direction of such vector using the known radius which is

$\frac{32}{\pi}$.

Horizontal:

$\frac{32}{\pi}\cos\left(\frac{5\pi}{3}\right)$

$\frac{32}{\pi}\cos\left(\frac{5\pi}{3}\right)$

$\frac{32}{\pi}\times\frac{1}{2}=\frac{16}{\pi}$

Vertical:

$\frac{32}{\pi}\sin\left(\frac{5\pi}{3}\right)$

$\frac{32}{\pi}\times\frac{-\sqrt{3}}{2}=\frac{-16\sqrt{3}}{\pi}$

Therefore the position vector would be:

$\frac{16}{\pi}\hat{i}-\frac{16\sqrt{3}}{\pi}\hat{j}$

But that's how far I went:

A factorization led me to:

$\frac{16}{\pi}\left( \hat{i}-\sqrt{3}\hat{j}\right )$

Therefore the acceleration would be:

$18\pi \times \frac{16}{\pi}\left( \hat{i}-\sqrt{3}\hat{j}\right )$

However this does not appear within the alternatives. Could it be that I'm not getting the right picture?

But since I don't know that value or the condition I don't know what further thing I can do. What about the other quantities given. How can I relate them? Can somebody help me here?.

Answer 1

Sketch of a solution:

The centripetal acceleration has a constant magnitude and it points to the center of the circle. So one way would be to calculate it's position az $t=4$s, then construct the vector with the given properties. Another one would be to calculate it's $\vec{r}(t)$ function, differentiate it twice and substitute in $t=4$s.

Answer 2

Since what is being asked is the position all that would be left to do is to take the sine and cosines functions to obtain the direction of such vector using the known radius which is

$\frac{32}{\pi}$.

That's where you made your first error. The radius has nothing to do with the direction of acceleration. You should have just stuck to sine and cosine.

You did correctly calculate the position vector of the sample at $t = 4.$ But the position vector is in the opposite direction from the acceleration vector (that's your second error; the position vector points from the center to the sample, but the centripetal acceleration points from the sample back toward the center), and again its magnitude is not part of "direction".

The magnitude of the position vector is the radius, which does affect the magnitude of acceleration, but you already had accounted for that effect when you computed $a_c.$

So in the end you have put two factors of $\frac{32}{\pi}$ into the result twice when there should only have been one. Take one factor of $\frac{32}{\pi}$ back out of your answer and you get $9\pi \times \left( \hat{i}-\sqrt{3}\,\hat{j}\right ).$ Next, remember that you used the exact opposite direction vector from the one you should have; reverse the direction of the vector and you have $9\pi \times \left( -\hat{i}+\sqrt{3}\,\hat{j}\right ).$