Question

A particle is moving in a circle of radius R in
such a way that at any instant the normal and
tangential component of its acceleration are
equal. If its speed at t=0 is v°. The time taken to
complete the first revolution is
​

Answer 1

For circular motion, the radial acceleration, `a_r` is given by `v^2/R` , and the tangential acceleration, `a_t` is given by `(dv)/(dt)` .

Since these accelerations are equal, so `v^2/R=(dv)/(dt)`

Hence, `(R/v^2)dv = dt`

Upon integration,

`int(R/v^2)dv = intdt`

`rArr –R/v=t+C`

To evaluate integration constant, C, put` v=v_o at, t=0`

Therefore, `C=-R/v_o`

The relation between v and t is,

`–R/v=t-R/v_o`

`rArr t=R[(v-v_o)/(v_o*v)]`

Now, `v=(ds)/(dt)` , where, s is the length of the arc covered by the particle as it moves in the circle.

Therefore, `t=R/v_o-R/((ds)/(dt))`

`=R/v_o-R*dt/(ds)`

`rArr tds=R/v_o-Rdt`

`rArr ds(R/v_o-t)=Rdt`

`rArr (ds)/R=dt/(R/v_o-t)`

Upon integrating,

`int(ds)/R=intdt/(R/v_o-t)`

`rArr s/R=-ln(R/v_o-t)+C`

putting at t=0, s=0

`C=lnR/v_o`

Therefore, the relation takes the form,

`s/R=lnR/v_o-ln(R/v_o-t)=ln(R/(R-v_ot))`

For a complete revolution, putting `s=2piR, t=T` (the time period),

`(2piR)/R= ln(R/(R-v_oT))`

`rArr T=R/v_o(1-e^(-2pi))`