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Having trouble understanding this.by mastermind3001
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#1
Mar805, 10:50 AM

P: 6

When dealing with centripetal force, you run into the formula [tex]a_c=v^2/r[/tex]. Where did that formula come from? I feel like knowing this will help me understand the concept better and not just the mechanics of it, but I can't figure it out. I think I'm just not in sync with mapping circular motion to linear motion.



#2
Mar805, 11:44 AM

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http://www.physicsforums.com/showthr...171#post482171 Also look here for the same derivation using polar coordinates: http://en.wikipedia.org/wiki/Centripetal_force AM 


#3
Mar805, 11:45 AM

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The way it is usually taught is:
Consider a particle in uniform circular motion. Then as the particle moves from P1 to P2 on the circle in a time [itex]\Delta t[/itex] the change in velocity [itex]\Delta v[/itex] can be drawn on a diagram. (Draw [itex]\vec v_1[/itex], [itex]\vec v_2[/itex] and [itex]\Delta \vec v[/itex], they form a triangle.) Since the motion is circular [itex]\vec v_1[/itex] and [itex]\vec v_2[/itex] are perpendicular to [itex]\vec r_1[/itex] and [itex]\vec r_2[/itex] and the angle between them is the same as the angle between and [itex]\vec v_1[/itex], [itex]\vec v_2[/itex]. You thus get similar triangles: [tex]\frac{\Delta \vec v}{v}=\frac{\Delta s}{R}[/tex] So that: [tex]a=\lim_{\Delta t \to 0}\frac{\Delta \vec v}{\Delta t}=\frac{v}{R}\frac{ds}{dt}=\frac{v^2}{R}[/tex] 


#4
Mar1105, 10:55 AM

P: 6

Having trouble understanding this.
I want to clarify something, because it's confusing the hell out of me. Velocity is a vector quantity, containing angle and magnitude. When working with anything linear, you usually work with the magnitude only (at least as far as I have done in my physics class). Here acceleration is the rate of change of the magnitude of the velocity. In circular motion, is the acceleration the change in the angle of the path of the object? Do you just not work with it's tangential velocity (magnitude) at all? It doesn't show up in [tex]a_c=\frac{v^2}{r}[/tex], which expanded is [tex]a_c=\frac{(\frac{2{\pi}r}{T})^2}{r}[/tex]. And if all this is right, then isn't the acceleration always constant? Someone please clarify. I just don't understand what acceleration is when it comes to circular motion.



#5
Mar1105, 11:32 AM

P: 599

The acceleration for centripetal motion can perhaps best be viewed componentswise. Look for a moment at only the xcomponent (say), ofcourse this changes over time yielding a nonzero xcomponent of the acceleration. 


#6
Mar1105, 11:45 AM

Mentor
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#7
Mar1405, 10:34 AM

P: 6

Is centripetal acceleration then just the change in the angle of the moving object per unit time?



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