Question

Let v and a be the instantaneous velocity and acceleration of a particle moving in a circle. Then what is the rate of change of speed dv/dt of the particle?

Answer 1

In  a circle the particle moves with a uniform speed | v |.

a = instantaneous acceleration = dv / dt

Let the center of circle be O. Let at t = t, the particle be at A (r , 0), where r = radius of the circle.   At time t = t +dt the particle moves to B by dθ radians.  Let  dv be the change in the velocity from v to v+dv.  The angle between these two vectors is dθ.

If we construct a triangle with vectors v and v+dv, then dv is the third side.  That is equal to v dθ = length of the arc in the infinitesimally small triangle.  Magnitude of the vector v+dv = | v | = constant speed.

dv = v dθ
dv/dt = v dθ/dt = v ω = r ω² = v²/r  = a ,            as  v = r ω