Hola there!!✔✔ ....derive that acceleration equals to = v^2/r...in case of centripetal force....
why is it so...!!??
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hey buddy,
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 2πr. 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 2πv.
The magnitude of acceleration is then
change in velocity/elapsed time,
which we can write as:
a = 2πv/(2πr/v) = v^2/r.
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 2πr. 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 2πv.
The magnitude of acceleration is then
change in velocity/elapsed time,
which we can write as:
a = 2πv/(2πr/v) = v^2/r.
joshihimanshu46:
ya
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