Circular motion (approximation of centripetal acc.)by V0ODO0CH1LD Tags: approximation, centripetal, circular, motion 

#1
May112, 06:41 PM

P: 201

Say I have a body moving in a circle of radius r with a constant velocity v.
The time it takes the body to go around the circumference once is: T = 2πr/v Then the time it takes the body to go around a fourth of the circumference is T/4. Now, imagine a diagram where when the body is at the leftmost portion of the circumference its velocity vector is pointing straight up, and when it is at the upmost portion of the circumference its velocity vector points entirely to the right. That is the body travelling over a fourth of the circumference. Now, the magnitude of the acceleration vector required to make the body behave that way over a period of T/4 is (v√2)/(T/4). And (v√2)/(T/4) is an approximation of the centripetal acceleration for this case. My question is: why is that an approximation? Why if I do the same thing over a smaller change in time I get closer to the centripetal acceleration? A circle is symmetric and the velocity is constant, right? 



#2
May112, 07:07 PM

Mentor
P: 11,985

[tex]\vec{a_{avg}}=\frac{\Delta \vec{v}}{\Delta t}[/tex] 



#3
May112, 07:30 PM

P: 201

That helped a lot! Thanks!
One for additional question though. How does all of that get summarised into (ω^2)*r or (v^2)/r? 



#4
May112, 08:09 PM

Mentor
P: 11,985

Circular motion (approximation of centripetal acc.)
You're welcome
This explains things pretty well: http://scienceblogs.com/dotphysics/2...acceleration/ 


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