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a particle is moving in a circle of radius R in such a way that at any instant the norml &tangential components of its accelkeration are equal if its speed at t=0 is v the time taken to complete the first revolution?
ω2r=αr, which implies that α=ω2,u=rω,where u is tangential velocity
u=v+αt,ω=(v+(alpha*t))/R,
ω=(v+ω2t)/R,now we have obtained a quadratic equation in omega.
Solve it to obtain omega as a function of 't'
from v=R*(omega),find v as a function of t,in addition,alpha=(omega)^2
u=v+(alpha*t)=v+ω2t
∫(v+αt)*dt (from t=0 to t=T)=2pi*R
you will get an equation in ' t',solve it to obtain the answer 'T'
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