A) \[\frac{R}{{{v}_{0}}}{{e}^{-2\pi }}\]
B) \[\frac{R}{{{v}_{0}}}(1-{{e}^{-2\pi }})\]
C) \[\frac{R}{{{v}_{0}}}\]
D) \[\frac{2R}{{{v}_{0}}}\]
Correct Answer: B
Solution :
[b] Total acceleration makes an angle of \[45{}^\circ \] with radius, i.e., tangential acceleration = radial acceleration or \[R\alpha =R{{\omega }^{2}}\] or \[\alpha ={{\omega }^{2}}\] or \[\frac{d\omega }{dt}={{\omega }^{2}},\] or \[\frac{d\omega }{{{\omega }^{2}}}=dt\] i.e. \[\int\limits_{{{\omega }_{0}}}^{\omega }{\frac{d\omega }{{{\omega }^{2}}}}=\int\limits_{0}^{t}{dt},\] or \[\omega =\frac{{{\omega }_{0}}}{1-{{\omega }_{0}}t}\] or \[\frac{d\theta }{dt}\frac{{{\omega }_{0}}}{1-{{\omega }_{0}}t},\] or \[\int\limits_{0}^{2\pi }{d\theta }=\int\limits_{0}^{t}{\frac{{{\omega }_{0}}dt}{1-{{\omega }_{0}}t}}\] i.e. \[t=\frac{1}{{{\omega }_{0}}}(1-{{e}^{-2\pi }})\] \[\left( {{\omega }_{0}}=\frac{{{v}_{0}}}{R} \right)\] or \[t=\frac{R}{{{v}_{0}}}(1-{{e}^{-2\pi }})\]
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