How do I find the average angular velocity between two points?

Jun 2017
345
6
Lima, Peru
The problem is as follows:

In the figure from below it is shown an observer who has put himself at the center of the coordinate system. He sees an object moving in a circular trajectory. If the average speed between $A$ and $B$ is $\left ( -2\hat{i}+\hat{j} \right )\frac{m}{s}$ and its position on point $A$ is $5\hat{i}\,m$. Find on $\frac{rad}{s}$ the average angular velocity between $A$ and $B$ if the time the object takes to get from $A$ to $B$ is $4\,s$.​



The alternatives on my book:

$\begin{array}{ll}
1.&0.35\hat{k}\\
2.&0.5\hat{k}\\
3.&0.55\hat{k}\\
4.&0.6\hat{k}\\
5.&0.75\hat{k}\\
\end{array}$

For this particular problem I'm stuck at how to use the information given the average velocity and the position. However I recall that when the word average is mentioned it mean this formula?

$\overline{v}=\frac{\vec{r}}{\Delta t}$

But other than that I'm not sure if it applies in this situation. Can somebody offer some help with this question?. :)
 
Last edited:

skeeter

Math Team
Jul 2011
3,276
1,769
Texas
$\Delta r = (-2i+j) \cdot 4 = -8i+4j$

$r(4)-r(0)=-8i+4j$

$r(4)-5i=-8i+4j \implies r(4)=-3i+4j$

$\theta (4) = \pi + \arctan\left(-\dfrac{4}{3}\right)$

$\theta (0) = 0$

$\bar{\omega} = \dfrac{\Delta \theta}{\Delta t} = \dfrac{\theta (4) - \theta (0)}{4} = \dfrac{\pi+\arctan\left(-\frac{4}{3}\right)}{4} \approx 0.55k$
 
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Jun 2017
345
6
Lima, Peru
Wait a second. When you subtract

$r(4)-r(0)$

wouldn't it be (for the first component).

This

$\left(-8-5\right)\hat{i}=-13\hat{i}$

Or could it be that I'm not understanding it right the subtraction of vectors?. Can you help me with this part because the rest was obvioud but it is there where I'm stuck!.:eek:
 
Last edited:
Jun 2017
345
6
Lima, Peru
Sorry I had to re read what you did. It turns out that we don't know in advance the position vector for the second point so that what you used is to build an equation for obtaining that vector and from then the angle.

But I'd like a disclaimer I mean some theory behind on what is the difference between these concepts and how should be used or what do they mean?

average velocity

average speed

average acceleration

and

just acceleration.

Are these terms interchangeable?.
 

skeeter

Math Team
Jul 2011
3,276
1,769
Texas
Wait a second. When you subtract

$r(4)-r(0)$

wouldn't it be (for the first component).

This

$\left(-8-5\right)\hat{i}=-13\hat{i}$
First of all, understand that change in position is $\Delta r = r(t_f) - r(t_0)$

You were given the average angular velocity and the time. We use these two pieces of info to determine the change in position in the x-y plane.

$\Delta r = \omega \cdot t = (-2i+j) \cdot 4 = -8i + 4j$

$r(0) = 5i + 0j$ at point A is given ... all you know about $r(4)$ at point B is what you can see in the diagram. It is located in quadrant II, hence a negative x-component and positive y-component.

Let $r(4) = ai + bj$ ... like I stated earlier, all we know now is $a<0$ and $b>0$.

$\Delta r = r(4) - r(0)$

$-8i + 4j = (ai + bj) - (5i + 0j)$

add $(5i+0j)$ to both sides ...

$(-8i+4j)+(5i+0j) = ai + bj$

$-3i + 4j = ai + bj \implies a = -3 \text{ and } b=4$ ... a vector in quad II with distance of 5 from the origin.

------------------------------------------------------------------------

Sorry I had to re read what you did. It turns out that we don't know in advance the position vector for the second point so that what you used is to build an equation for obtaining that vector and from then the angle.

But I'd like a disclaimer I mean some theory behind on what is the difference between these concepts and how should be used or what do they mean?

average velocity

average speed

average acceleration

and

just acceleration.

Are these terms interchangeable?.
Obviously, this venue is not conducive to fully explaining the definition of terms with regard to circular motion and their nuances. Here is a link to a source that covers the basics ...

Angular Motion, General Notes