The problem is as follows:
A spheric strawberry candy is spinning inside a centrifuge in a confectionery. Find the magnitude of the acceleration in $\frac{m}{s^{2}}$ of the candy after $4\,s$ from the beginning of its rotation. It is known that the arc length follows the function $s(t)=1+t^3$ where $t$ is the elapsed time. It is also known that after $2\,s$ from its rotation the normal (centripetal) acceleration is $2\,\frac{m}{s^{2}}$.
The given alternatives in my book are as follows:
$\begin{array}{ll} 1.&40\,\frac{m}{s^{2}}\\ 2.&35\,\frac{m}{s^{2}}\\ 3.&30\,\frac{m}{s^{2}}\\ 4.&45\,\frac{m}{s^{2}}\\ 5.&50\,\frac{m}{s^{2}}\\ \end{array}$
I attempted to solve this problem and the steps from below describe my effort however I came stuck at the end as I couldn't really find a way to relate the acceleration which is what is being asked to what the arc lenght is.
The only think which I can recall here is that the acceleration is the second derivative of the length. Since the function over time is given then that would be the tantential acceleration.
Since
$s(t)=1+t^3$
Then
$s'(t)=3t^2$
and
$s''(t)=6t$
So by evaluating this function on t=4 gives me the tangential acceleration at that given instant.
$s''(4)=6(4)=24$
But then what?.
I can relate that the centripetal acceleration (which is what is being asked) is given by:
$a_{c}=\omega^2 r$
However the angular speed to be used here is that attained after $4$ seconds have elapsed.
But the problem is to find the radius.
But we don't know that:
But we do know what is the value of the arc length.
$s(2)=1+2^3=9\,m$
and the centripetal acceleration for that given time:
$a_{c}=\omega^2 r = 2$
So what we need is the angular speed at $2\,s$ as a function of the radius.
Since what we need here is the angular acceleration, we can relate it with the tangential acceleration by the equation:
$a_{t}=6t=6\times 2 = 12$
$\alpha=\frac{a_{t}}{r}=\frac{12}{r}$
Therefore we have the angular acceleration in terms of the radius.
Thus:
$\omega_{f}=\omega_{o}+\alpha t$
$\omega_{f}=0+\frac{12}{r}\times 2 = \frac{24}{r}$
Then we have the angular speed as a function of $r$.
This can be replaced in the given value for the centripetal acceleration.
$a_{c}=\omega^2 r = 2$
$\left(\frac{24}{r}\right)^2r = 2$
Therefore:
$r=288\,m$
Then using this value we can compute the centripetal acceleration:
But we do need the angular acceleration for $t=4$:
$\alpha=\frac{a_{t}}{r}=\frac{24}{288}= \frac{1}{12}$
I noted that this is different than the angular acceleration at $t=2$
$\alpha=\frac{12}{288}= \frac{1}{24}$
Now arises the question. What sort of "acceleration" is the problem asking?
Perhaps is the magnitude of the total acceleration or the norm that is happening because the tangential acceleration and the centripetal acceleration?
If so that would be:
$a=\sqrt{\left(\frac{1}{12}\right)^{2}+\left(24\right)^2}$
Then this would become into:
$a\approx 24$
But this doesn't appear in any of the alternatives given. As I mentioned I'm still confused at what was the intended meaning in this question with the word acceleration, did the person who posed this problem referred to tangential acceleration or a total acceleration in other words the norm of both the tangential and centripetal acceleration?
Can somebody help me here?. What's exactly the part where I got lost?.
You have two errors. First, the acceleration (and therefore the angular acceleration) is not constant. So the formula for the $\omega_f$ is wrong. Why don't you just use $\omega (t)=\frac{s'(t)}r$? The second problem is that you try to add the angular acceleration with a centripetal acceleration. One of them has units of radians/s$^2$, the other one is m/s$^2$.
Here is what I did: $$a_c(t)=\omega^2(t)r=\frac{(s'(t))^2}r$$ Then $$\frac{a_c(4)}{a_c(2)}=\left(\frac{s'(4)}{s'(2)}\right)^2=\left(\frac{3\cdot 4^2}{3\cdot 2^2}\right)^2=16$$ so $a_c(4)=32$. With the tangential acceleration $a_t(4)=24$, I get $$a=\sqrt{32^2+24^2}=40$$