Physics teacher wanted me to understand how to derive

In summary, the conversation discusses the process of deriving the equation a=v^2/r using the equations v=d/t, \omega=\theta/t, and arc length=r\theta. The conversation includes hints, tips, and discussions on how to approach the derivation, including using calculus and geometry. Ultimately, the conversation concludes that the "trick" to finding the solution is to find the difference in the velocity vectors between two points on the circle separated by \Delta\theta. The conversation also mentions the importance of understanding vectors and the potential career opportunities in physics that can come from understanding the derivation process.
  • #1
jimmy p
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My Physics teacher wanted me to understand how to derive [tex]a=v^2/r[/tex] using the equations [tex]v=d/t[/tex], [tex]\omega=\theta/t[/tex] and [tex]arc length=r\theta[/tex] but when it came down to it, i had brain freeze..and when my teacher looked at it, he had brain freeze. This isn't part of my syllabus but i was wondering how you derive the equation (so it isn't homework!)

oh yeah, i was given the hint to work it first into the equation [tex]v=r\omega[/tex] and then work from there.

Cool, i just used that LaTex stuff! that was quite challenging!

thanx

Jimmy P
 
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  • #3
OK unfortuantely i didnt find that too useful. That is kind of what we tried to do but it failed miserably, because we ended up getting T on both sides of the equation. i could get to [tex]v=r\omega[/tex] (if that is even helpful) and couldn't get any further
 
  • #4
Can you use calculus?
 
  • #5
Originally posted by jimmy p
OK unfortuantely i didnt find that too useful. That is kind of what we tried to do but it failed miserably, because we ended up getting T on both sides of the equation. i could get to [tex]v=r\omega[/tex] (if that is even helpful) and couldn't get any further
Well, those two threads show how to derive the formula for centripetal acceleration. I know of no other way.

Using the equations you started with, you can certainly get to [tex]v=r\omega[/tex]. But you won't get any further without using the strategy explained in those threads. The "trick" is to find [tex]\Delta V[/tex]---the difference in the velocity vectors---between two points on the circle separated by [tex]\Delta\theta[/tex].

Why don't you show what you've done and perhaps we can fix it.
 
  • #6
You have everything EXCEPT a formula for [tex]\inline{\Delta v}[/tex]. This is the real physics; you have to think about it. The rest is just formula manipulation. I remember when I was in high school, I thought about this very thing for days, because I had difficulty coming to terms with how vectors add; I did not understand [tex]\inline{a=\Delta v/\Delta t}[/tex] as a VECTOR equation. But I put in the effort and was rewarded with a career in physics.
 
  • #7
I haven't looked at either of those links, but I remember first deriving the formula by creating a proportion between the triangle the velocity and acceleration vectors created, and the triangle that the 2 radii and the chord created.
 
  • #8
cant anyone actually give the answer? i promise that it isn't homework and that i have tried!
 
  • #9
Try this.
Draw a vector for r with a vector for v on the end (will be 90 degrees from this), then a little time later examine what would happen. The arclength s= r theta is an approximation from geometry.

I haven't done this yet myself, but i am working on it
 
  • #10
for the algebraic way you are looking for you go from v = d/t and find d from the other eqns and get v = r omega
Then with acceleration, a = v / t, substitue the v and you get r omega / t, v=r/t so a = v omega, the trick is a = v^2 omega / v = v^2 omega / ( r omega ) and cancel.

BUT this gives you nothing (not exactly true). Calculus method and geometry - what i siad to do in last post, work the same and give you a good in sight into what happend.
This is related to how Feynman got to grips with QED.
 
  • #11
Although with the geometrical method i can't avoid the final 'trick' bit to go from a = r omega^ 2 to v^2 / r
Doh
 
  • #12
Originally posted by jimmy p
cant anyone actually give the answer?
We have. Several times over. You just don't seem to want to accept it! :smile:
 
  • #13
lol sorry, i wasnt concentrating the other night when i wrote that, thanks guys!
 

1. How do I derive equations in physics?

Deriving equations in physics involves using mathematical principles and reasoning to arrive at a formula or relationship between variables. This is typically done by starting with known equations and using algebraic manipulations and scientific principles to rearrange the terms and solve for the desired variable.

2. What is the purpose of deriving equations in physics?

The purpose of deriving equations in physics is to gain a deeper understanding of the underlying principles and relationships between different physical quantities. It allows us to predict and explain the behavior of physical systems and make accurate calculations.

3. Can I derive equations on my own or do I need to follow a specific method?

While there are general guidelines and techniques for deriving equations in physics, there is no set method that must be followed. As long as you have a strong understanding of the relevant concepts and are able to use mathematical reasoning, you can derive equations on your own.

4. What are some common mistakes to avoid when deriving equations in physics?

One common mistake is forgetting to take into account all relevant physical quantities or assumptions. It is important to carefully consider all variables and conditions when deriving equations. Another mistake is making algebraic errors or skipping steps, which can lead to incorrect equations.

5. How can I improve my skills in deriving equations in physics?

The best way to improve in deriving equations in physics is to practice regularly and seek feedback from others. You can also review and understand the derivation process behind well-known equations and apply similar techniques in your own derivations. Additionally, developing a strong understanding of mathematical principles and physical concepts will greatly aid in deriving equations in physics.

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