Centripetal force formula technique

Today our topic is about derivation of centripetal force in previous post we have use this force formula but in this topic you will learn short derivation technique for centripetal force rather than traditional derivation in different books i hope you will enjoy to learn derivation of centripetal force before start this derivation i will suggest you to go through previous post centripetal and centrifugal force concept so that you can understand better concept.

Centripetal force derivation by two method

we will derive formula for centripetal force by two different method one we called traditional method and other is short technique method first we will see traditional method for centripetal force formula derivation and then for short method you should know both the method for depth concepts for centripetal force derivation.

# 1 centripetal force formula derivation by traditional method

From the above picture let V1 and V2 is velocity of particle at two different time now we assume that particle is doing uniform circular motion its mean that magnitude of velocity or speed is constant | V1 | = | V2 | = V =constant.
only its velocity is changing due to change in direction, Here in circular motion every time velocity direction is changing hence there is change in velocity and we know that due to change in velocity acceleration is created and we have already learned in previous post  due to change in direction an acceleration is generated which is centripetal acceleration hence from Newton’s second law centripetal force is acting in circular motion its direction is towards the center of the circle along the radius.
since velocity is always tangent to circular path means angle between radius and velocity V1 and V2 is always 90 degree angle between r and r is ϴ for large value of ϴ triage is given where angle between V1 and V2 is ϴ but how it is simple we see that in circle two tangent  V1 and V2  the perpendicular radius making an angle ϴ  hence its tangent line angle will be also ϴ since magnitude of V1 and V2 is same so its length will be equal in triangle hence from from triangle apply vector we can write.
→     →      →        →      →    →
V1 + ∆V  = V2  or  ∆V  = V2 – V1  this is change in velocity which create acceleration called centripetal acceleration.
now we say this is for very less time dt then angle ϴ will be also very small and taken as dϴ in reality then velocity change will also be very small hence  ∆V  = dv.
now for dt time linear distance will also be very small ds hence taken other triangle ds now dv and ds triangle are similar because all three side of both triangle will be equal for very small value of dϴ  hence ration of side will be same so from here we can write as

dv/v = ds/r  now divide by dt on both side we will get note v1 = v2 = v
dv/v*dt  = ds/r*dt  or  (dv/dt)*1/v  = (ds/dt)*1/r   ( dv/dt = a?  = centripetal acceleration) ds/dt =v in above equation

a?/v  =  v/r  or a? = v²/r  or  a? = v²/r now we can also put v = ?r then we will get

a?   = ?²r²/r = ?²r now for centripetal force we can write F? = ma? = mv²/r

F?  = m?²r this is centripetal acceleration.

Centripetal force very important point why centripetal acceleration direction is towards center of the circle ?

Ans if centripetal acceleration and velocity is perpendicular then take component of centripetal acceleration along the velocity then it will be a?cos90 which will be equal to zero then this acceleration will not increase or decrease the speed of particle hence it will be uniform circular motion. if centripetal acceleration not perpendicular to velocity the speed of particle will not be constant which will not be uniform circular motion proved.

# 2 centripetal force formula derivation technique

Assumption Speed constant | v | = constant for uniform circular motion in uniform circular motion only centripetal acceleration no tangential acceleration only velocity direction change no change of magnitude of velocity.

we have already studies in previous post v =  ?r  if you have not seen this post you must read that post because we have discussed in details about this relation your concept of circle motion will be complete clear all doubt please refer this post

now come to our equation v =  ?r  here v velocity will not change r will also not change hence ? will also not change so it is very important in uniform circular motion v is constant as well as ? is also constant note it .

v =  ?r  now differentiate this equation with respect to time on both side see what we get
dv/dt = ?(dr)/dt +r(d?)/dt  = ?(dr)/dt +0  here ? is constant (using by parts method)

now here velocity can change by two method either by changing by magnitude or by changing by direction so above we have already made assumption velocity magnitude is constant hence only direction is change here and we know due to change in direction centripetal acceleration is created hence dv/dt = a? it is clear

dv/dt  = ?(dr)/dt  ( here dr/dt is linear velocity v )

a?  = ?v  ( now put the value v =  ?r from above in this equation )

a? = ?*?r = ?²r  or centripetal force  F? = ma? = m?²r  = mv²/r  hence it is again proved

F? = ma? = m?²r  = mv²/r  centripetal force formula .
i will suggest you to refer method first for your 11th exam but for competition you can refer method second.