12.05.2024

Hello in this video I'm going to derive. the formula for centripetal acceleration. v square over R let's have an object. moving along a circular arc of radius R. it underwent angular displacement of. theta and let's call the velocity at. this and this position v1 and v2 let's. draw a vector triangle involving v1 and. v2 now what does this side of the. triangle represent it represents the. change in velocity Delta V see the. origin of velocity plus the change in. velocity is equal to the final velocity. do you see that this angle here is. actually the same angle people here see. R and R forms the angle theta so v1 and. v2 should also found the angle theta. because V 1 and V 2 adjust these two R's. turn 90 degrees let's draw another. triangle based on R and R of course this. angle here would be theta so what should.

That side of this triangle correspond to. now look here. the arc length this act length should be. exactly equal to V delta T so delta T is. the amount of time taken for the the. object to go from here to here and V is. just a speed V is the same as V y is the. same as V 2 because this guy is going to. constant speed so the act length is V. delta T if theta is very small then the. arc length is practically the same. length as the line joining these two. positions get it. the curvy act length and the straight. line joining these two points are. practically the same if the pay is small. enough so the third side of the triangle. here is basically equals to the arc. length which is V delta T so some. students are going to say hey are we. cheating because the two lengths are not. exactly the same right not true because.

Um we are trying to work out the. instantaneous acceleration so the. angular displacement is really very very. small and when the angle is very very. small. the two laps are really practically the. same for example you go to the swimming. pool you look at a water surface is it a. straight line yeah it looks straight but. you know that it can be a straight line. because the earth is round so that. straight line that you thought with a. straight line is actually part of an arc. so we are really making a very very. approximation here now back to the two. triangles do you realize that these two. are similar triangles because they are. both isosceles triangles subtended by. the same angle theta can we do if. similar triangles we can equate the. ratio of the corresponding sides right. so we can write Delta P over V is V 1 or.

V 2 but they are the same half so we. just call it V so Delta V divided by V 1. should be equal to the ratio of the. corresponding sides on the other. triangle which is a V delta T divided by. our rearrange the equation you get Delta. V over the other piece because v square. over r when delta T is small enough. Delta V over delta T is actually the. instantaneous acceleration which is v. square over R so with some simple. geometry we have shown that when an. object is doing circular motion at a. constant speed v radius R then it's. velocity is changing at the rate of risk. level R that's all Tata