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_o#, the time taken to complete the first revolution is?

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Answer 1 · 2017-12-22T10:13:06

#t = R/v_0(1 - e^(-2pi))#

Calling #hat n = (cos theta, sin theta)# and #hat tau = (-sintheta, costheta)# we have

#p = R hat n rArr dot p = R dot theta hat tau rArr ddot p = R ddot theta hat tau - R (dot theta)^2 hat n#

We have

#R ddot theta = R (dot theta)^2 rArr dot v_t = 1/R v_t^2#

Here #v_t = # tangential speed.

Integrating

#1/v_t +t/R = 1/v_0# but #v_t = (ds)/dt# so

#(ds)/(dt) = (R v_0)/(R-t v_0)# and integrating again

#s = s_0 - R log_e((R-t v_0)/R)# but #s_0 = 0# and

#2piR = -R log_e((R-t v_0)/R)# then

#t = R/v_0(1 - e^(-2pi))#

Answer 2 · 2017-12-22T10:31:15

Continuing with your approach:

Given that the particle starts with the velocity #v_0# to move in a circular path of radius R. It acquires tangential velocity #v# at t th instant.

It is also given that at any instant the normal and tangential component of its acceleration are equal.
So we can write

#v^2/R = v(dv)/(ds)#,where #s# represents its path length.
#(ds)/R =(dv)/v#

Now at #t=0,s=0 and v =v_0#

at #t =T" the time of first revolution "#

#s=2piR and v=V#

#int_0^(2piR)(ds)/R = int_(v_o)^V(dv)/v#

#=>2pi = lnV - lnv_o=ln(V/v_0)#

#=>v_oe^(2pi) = V#

#=>v_0/V=e^(-2pi)......(1)#

Applying same condition we can also write

#v^2/R=(dv)/(dt)#

#=>int_0^T(dt)/R=int_(v_0)^V(dv)/v^2#

#=>T/R=1/v_0-1/V#

#=>T=R/v_0(1-v_0/V).....(2)#

Combining (1) and (2) we get

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