Friday, November 16, 2018

homework and exercises - A simple derivation of the Centripetal Acceleration Formula?


Could someone show me a simple and intuitive derivation of the Centripetal Acceleration Formula $a=v^2/r$, preferably one that does not involve calculus or advanced trigonometry?



Answer




Imagine a object steadily traversing a circle of radius $r$ centered on the origin. It's position can be represented by a vector of constant length that changes angle. The total distance covered in one cycle is $2\pi r$. This is also the accumulated amount by which position has changed..


Now consider the velocity vector of this object: it can also be represented by a vector of constant length that steadily changes direction. This vector has length $v$, so the accumulated change in velocity is $2 \pi v$.


The magnitude of acceleration is then $\frac{\text{change in velocity}}{\text{elapsed time}}$, which we can write as: $$a = \frac{2 \pi v}{\left(\frac{2\pi r}{v} \right)} = \frac{v^2}{r} \,.$$


Q.E.D.




Aside: that derivation is used in a lot of algebra/trig based textbooks.


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