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November 3rd, 2019, 05:19 PM   #1
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Question How can I find the tangential speed of an object from vectors?

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.


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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November 3rd, 2019, 06:01 PM   #2
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$a_c = \dfrac{v_T^2}{r} \implies v_T = \sqrt{r \cdot a_c}$

also, note $a_T = 50 \, m/s^2$ is opposite in direction to $v_T$
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November 4th, 2019, 03:16 AM   #3
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Quote:
Originally Posted by skeeter View Post
$a_c = \dfrac{v_T^2}{r} \implies v_T = \sqrt{r \cdot a_c}$

also, note $a_T = 50 \, m/s^2$ is opposite in direction to $v_T$
Therefore the sample is decelerating and by doing the computation that would be $10\frac{m}{s}$
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