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The problem is as follows:

A protein sample spins in the counterclockwise direction in a centrifuge seen from the top as shown in the diagram from below. The radius of the centrifuge es $R=2\,m$. The magnitude of its speed changes. At a certain instant the acceleration vector is as shown in the figure. Find the speed in $\frac{m}{s}$ and state the type of its motion in the given instant. A for acceleration if the speed increases or D for deceleration if the speed decreases.

Sketch of the problem

The alternatives given on my book are:

$\begin{array}{ll} 1.10\frac{m}{s};\,A\\ 2.10\frac{m}{s};\,D\\ 3.5\frac{m}{s};\,A\\ 4.5\frac{m}{s};\,D\\ 5.10\frac{m}{s};\,\textrm{uniform motion}\\ \end{array}$

This problem I'm particulary lost at. The acceleration shown in the graph. What is it?. Is it perhaps the total acceleration?. In other words the norm of the centripetal and the tangential acceleration?

If so then that would meant that the:

$a_c=50\frac{m}{s^2}$

Then the tangential acceleration will be:

$a_t=50\frac{m}{s^2}$

And because the angular acceleration is related to the tangential acceleration due the radius as:

$a_t=\alpha\times r$

Then:

$\alpha=\frac{a_t}{r}=\frac{50}{2}=25\frac{rad}{s^2}$

But that's how far I went in my analysis. What else can I relate to find the asked speed?.

The only equations which I recall are:

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

Can somebody help me here?. What exactly should be the right path to get the answer?.

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2 Answers 2

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The tangential acceleration has opposite direction from the tangential velocity so there will be tangential deceleration D.

The centripetal acceleration is $50 \frac{m}{s^2}$; as @guest stated, this leads to a velocity of $\sqrt{50 \frac{m}{s^2} \cdot 2m} = 10 \frac{m}{s}$.

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  • $\begingroup$ I understand that I could ask for a theoretical clarification but do you know by chance if there is any sort of relationship between centripetal acceleration and tangential acceleration does it exist an equation to relate one and the other?. $\endgroup$ Nov 4, 2019 at 2:49
  • $\begingroup$ No. They are completely independent. However, $\vec{a} = \vec{a}_c + \vec{a}_t$. $\endgroup$ Nov 4, 2019 at 8:27
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You have to decompose the acceleration vector into two components, one which is tangential to the orbit and the other which points towards the center of the circle. Note that the tangential component is paralel to the velocity $\vec{v}$ but it points in the opposite direction , so there is gonna be deceleration. The component pointing towards the center is the centripetal acceleration given by $$a_{cp} = \frac{v^{2}}{R} $$. Remember each component of the acceleration can be calculated using pythagoras theorem.

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  • $\begingroup$ Gee. I totally overlooked that, but that's right. One vector is pointing in the opposite from the speed. However does it exist an equation which can relate the tangential acceleration and the centripetal acceleration?. Or one is totally different from the other?. $\endgroup$ Nov 4, 2019 at 2:48
  • $\begingroup$ They are completely independent. A centripetal acceleration is needed to keep the particle in a circular trajectory. It has nothing to do with the tangential motion: the particle could be accelerating tangentially or not. However, note that the general formula for the centripetal acceleration is $a_{cp}=v^{2}/R$ so if the particle is supposed to keep its distance from the center of the circle, then an increase of $v$ should imply an increase on $a_{cp}$ and, as a consequence, an increase on the centripetal force $F_{cp}=ma_{cp}$. $\endgroup$
    – IamWill
    Nov 4, 2019 at 11:50

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