A particle in a circular path speeds up with a uniform rate between two diametrically opposite points of a circle of radius R. If its time of motion between these two points is equal to T, find the accelertaion of the particle averaged over the time T.

A particle in a circular path speeds up with a uniform rate between tw

1.	A particle in a circular path speeds up with a uniform rate between two diametrically opposite points of a circle of radius R. If its time of motion between these two points is equal to T, find the accelertaion of the particle averaged over the time T.
Answer» Solution :The average acceleration of the particle when it revolves HALF of the circle is : `a_(a v) = (\|vec v - vec v_0\|)/(DELTA T)` where `vec v = -v hat j, vec v_0= v_0 hat j`, and `Delta T = T` (given). Then `a_(a v) = (v + v_0)/(T)` Let US express `v + v_0` in terms of `R and T`. As the particle moves with constant angular acceleration its average angular velocity is `omega_(a v) = (Delta theta)/(Delta t) = (omega + omega_0)/(2)` where `omega = (v)/(R). omega_0 = (v_0)/( R), Delta theta = pi, and Delta t = T` Then, `v + v_0 = (2 pi R)/(T)` Sunstituting the value of `v + v_0` from Eq. (II) in Eq. (i), we have `a_(a v) = (2 P)/(T^2)`. .