My Math Forum  

Go Back   My Math Forum > Science Forums > Physics

Physics Physics Forum


Thanks Tree1Thanks
  • 1 Post By skeeter
Reply
 
LinkBack Thread Tools Display Modes
October 27th, 2019, 03:48 AM   #1
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Question How do I find the average angular velocity between two points?

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 by Chemist116; October 27th, 2019 at 03:51 AM. Reason: missing characters
Chemist116 is offline  
 
October 27th, 2019, 04:59 AM   #2
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,101
Thanks: 1677

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

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

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

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

$\theta (0) = 0$

$\bar{\omega} = \dfrac{\Delta \theta}{\Delta t} = \dfrac{\theta (4) - \theta (0)}{4} = \dfrac{\pi+\arctan\left(-\frac{4}{3}\right)}{4} \approx 0.55k$
Thanks from topsquark
skeeter is offline  
October 27th, 2019, 02:17 PM   #3
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Question

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!.

Last edited by Chemist116; October 27th, 2019 at 02:19 PM. Reason: fixed characters
Chemist116 is offline  
October 27th, 2019, 02:24 PM   #4
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
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?.
Chemist116 is offline  
October 27th, 2019, 06:41 PM   #5
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,101
Thanks: 1677

Quote:
Originally Posted by Chemist116 View Post
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.

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

Quote:
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
skeeter is offline  
Reply

  My Math Forum > Science Forums > Physics

Tags
angular, average, find, points, velocity



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How Angular Velocity of Earth is 2π/T? Ganesh Ujwal Physics 1 July 10th, 2019 12:59 PM
Angular velocity aaronmooy47 Physics 6 February 2nd, 2017 09:15 PM
angular velocity unistu Algebra 2 January 7th, 2015 03:09 AM
Obtaining angular velocity given linear velocity and center of rotation 3D quarkz Calculus 0 April 18th, 2014 06:34 AM
Need an Algorithm (might be Angular Velocity Type) lothia Algebra 0 April 9th, 2009 02:27 PM





Copyright © 2019 My Math Forum. All rights reserved.