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 November 3rd, 2019, 04:01 PM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus How do I find the centripetal acceleration when vectors are given? 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 it 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 $\omega_{0}=\frac{\pi}{12}$ $\theta (t)=\theta_{o}+ \frac{\pi}{12} t + \frac{1}{2}\left(\frac{\pi}{6}\right) t^2$ $\theta (t)=\theta_{o}+ \frac{\pi}{12} t + \frac{\pi}{12}t^2$ $\theta (0)=\theta_{o}+ \frac{\pi}{12} (0) + \frac{\pi}{12}(0)^2$ $\theta(0)=\theta_{0}$ 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?.
 November 3rd, 2019, 04:35 PM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 $\omega = \omega_0 + \alpha t$ If the radius is constant, then $a_{centripetal} = \omega^2 r$.
November 3rd, 2019, 05:25 PM   #3
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Quote:
 Originally Posted by DarnItJimImAnEngineer $\omega = \omega_0 + \alpha t$ If the radius is constant, then $a_{centripetal} = \omega^2 r$.
Sorry but I still don't get it. How can I relate the centripetal acceleration with the angular acceleration which is what is being asked. My major source of confusion is what should I replace there? Do you imply the final angular speed?.

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

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

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

From there:

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

$a_{c}=18\pi$

But that's how far I went needless to say, how about the vectors shown in the alternatives?.

Does it imply that I must obtain the position?

From which should I use the tangent function to obtain the coordinates?.

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

$A\,\theta(0)=0$

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

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

$\theta_{0}=0$

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?

Last edited by Chemist116; November 3rd, 2019 at 06:18 PM.

 November 3rd, 2019, 05:56 PM #4 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle What does power ranger hold in his hand? It sure looks like a "cylinder with cap".
November 3rd, 2019, 06:34 PM   #5
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Quote:
 Originally Posted by tahirimanov19 What does power ranger hold in his hand? It sure looks like a "cylinder with cap".
He's not a power ranger. He's a sentai blue. More specifically Blue buster and what he is holding is the Ichigan buster which if you read the article (in the given links) is a camera which can transform into a gun when he's in battle mode and return to being a spy camera while on operations mode.

Okay going back to my question, can you perhaps take a look into it. Did I misunderstood any of the steps?.

 November 3rd, 2019, 07:19 PM #6 Math Team     Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 Is $r=32\pi$, or $\dfrac{32}{\pi}$ ? The original problem statement says $32\pi$, yet your calculations use $\dfrac{32}{\pi}$ I'll assume $r=\dfrac{32}{\pi}$ $\theta(4) = 0 + \dfrac{\pi}{12} \cdot 4 + \dfrac{\pi}{12} \cdot 16 = \dfrac{5\pi}{3}$ $\omega(4) = \dfrac{\pi}{12} + \dfrac{\pi}{6} \cdot 4 = \dfrac{3\pi}{4}$ $|a_c| = r \omega^2 = \dfrac{32}{\pi} \left(\dfrac{3\pi}{4}\right)^2 = 18\pi$ at $t=4$, the particle is at position $\theta = \dfrac{5\pi}{4}$, indicating the direction of $a_c$ (toward the center) is $\dfrac{2\pi}{3} = -\dfrac{1}{2} \hat{i} + \dfrac{\sqrt{3}}{2} \hat{j}$ $a_c = 18\pi\left(-\dfrac{1}{2} \hat{i} + \dfrac{\sqrt{3}}{2} \hat{j}\right) = 9\pi\left(-\hat{i}+\sqrt{3} \hat{j} \right)$ Thanks from Chemist116

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