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 November 4th, 2019, 02:41 AM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus How can I find the modulus of the angular acceleration? The problem is as follows: A pulley starts spinning from rest a rotation with constant angular acceleration. After $5\,s$ a point in its periphery has an instant acceleration which makes a $53^{\circ}$ angle with its linear speed. Find the modulus of the angular acceleration (in $\frac{rad}{s^{2}}$) of the pulley.The given alternatives in my book are as follows: $\begin{array}{ll} 1.&0.5\,\frac{m}{s}\\ 2.&0.53\,\frac{m}{s}\\ 3.&0.053\,\frac{m}{s}\\ 4.&0.106\,\frac{m}{s}\\ 5.&1.06\,\frac{m}{s}\\ \end{array}$ For this particular problem. I'm lost as how should I use the given information of the instant acceleration and the linear speed. How should I put those vectors?. Which sort of equation should I use?. The only equation which comes to my mind for the angular acceleration is how it is related to the tangential acceleration as: $a_{t}=\alpha \times r$ But in this case there is no radius. Thus I believe it has something to do with vectors but I can't really find exactly how to use that information. Can somebody help me here?.
 November 4th, 2019, 06:35 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 The thing to bear in mind here is that you have two components of the acceleration, $\vec{a} = \alpha r \hat{e}_t - \omega^2 r \hat{e}_r$ (plus two more terms that are zero if the radius is constant). In this problem, you don't know the radius, but you do know the angle (and therefore the ratio) between the tangential and centripetal accelerations. (Note $\hat{e}_r$ is centrifugal acceleration, away from the centre, not that it matters in this case.) Then replace $\omega$ with $\alpha t$. This gives you everything you need. BTW, either you or the book made a mistake with the units in the answers. They should be in $\frac{rad}{s^2}$.
November 5th, 2019, 01:12 AM   #3
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Quote:
 Originally Posted by DarnItJimImAnEngineer The thing to bear in mind here is that you have two components of the acceleration, $\vec{a} = \alpha r \hat{e}_t - \omega^2 r \hat{e}_r$ (plus two more terms that are zero if the radius is constant). In this problem, you don't know the radius, but you do know the angle (and therefore the ratio) between the tangential and centripetal accelerations. (Note $\hat{e}_r$ is centrifugal acceleration, away from the centre, not that it matters in this case.) Then replace $\omega$ with $\alpha t$. This gives you everything you need. BTW, either you or the book made a mistake with the units in the answers. They should be in $\frac{rad}{s^2}$.
Yes. I'm sorry for that. I wrote this question in a rush. It should had been as you correctly indicated. In other words as $\frac{rad}{s^{2}}$.

I'm confused at why the acceleration at the given instant is it a subtraction from these two accelerations, the tangential from the centripetal?

Wouldn't it be the sum?. Or could I be looking at those vectors differently from you? Or what it was intended to say with the negative sign is that just the centripetal is pointing inwards?.

I'm posting a sketch from what I'm understanding by reading your words. Please tell me if I did missunderstood what you meant.

If solely guided by what you mentioned then the tangent function can be used to relate both.

$\tan 53^{\circ}=\frac{\omega ^ 2 r}{\alpha\times r}$

$\frac{4}{3}=\frac{\omega ^ 2}{\alpha}$

Thus:

$\alpha=\frac{3 \omega ^2}{4}$

But I'm not given $\omega$:

Since I calculated it from rest:

$\omega_{f}=\omega_{0}+\alpha \left(5\right)$

$\omega_{f}=5\alpha$

By introducing this in the above equation:

$\alpha=\frac{3 \left(5\alpha\right) ^2}{4}$

Then:

$\alpha=\frac{4}{25 \times 3}\approx 0.0533$

But in this situation I'm not considering the sign you mentioned. Although by comparing this answer to the answers sheet in my book it checks.

But again I'd like you could check if what I'm understanding is correct.

 November 5th, 2019, 06:36 AM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 It is the vector sum. The negative sign, just as you said, is because the $\hat{e}_r$ unit vector points (by convention) away from the centre, while the acceleration is towards the centre. Thanks from Chemist116
November 7th, 2019, 12:47 AM   #5
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Quote:
 Originally Posted by DarnItJimImAnEngineer It is the vector sum. The negative sign, just as you said, is because the $\hat{e}_r$ unit vector points (by convention) away from the centre, while the acceleration is towards the centre.
So is my drawing or the way how I understood what you meant correct?.

November 7th, 2019, 06:12 AM   #6
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reference the diagram ...

$a:a_c:a_T = 5:4:3$

$\omega = \alpha \cdot t \implies \omega(5) = 5\alpha$

$a_c = r \cdot \omega^2 \implies a_c(5) = r \cdot 25 \alpha^2$

$a_T = r \cdot \alpha$, a constant

$\dfrac{a_c}{a_T} = \dfrac{4}{3} = \dfrac{r \cdot 25\alpha^2}{r \cdot \alpha} = 25 \alpha$

$\alpha = \dfrac{4}{75} \approx 0.053 \, rad/s^2$
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