A particle starts from rest and moves along a circle with a speed $v = ct$, where $c$ is a positive

Question

A particle starts from rest and moves along a circle with a speed $v = ct$, where $c$ is a positive constant. At the moment it completes exactly $N$ revolutions, the magnitude of its total acceleration is given by $c\sqrt{1 + k\pi^2 N^2}$. Find the value of the integer $k$.

Professor PC Thomas Classes · Answer

Tangential acceleration $a_t = dv/dt = c$. Distance $s = N(2\pi R)$. Also $s = \int v dt = \frac{1}{2}ct^2 \Rightarrow t^2 = \frac{4\pi N R}{c}$. Normal acceleration $a_n = \frac{v^2}{R} = \frac{c^2t^2}{R} = \frac{c^2}{R} \left(\frac{4\pi N R}{c}\right) = 4\pi N c$. Total acceleration $a_{tot} = \sqrt{a_t^2 + a_n^2} = \sqrt{c^2 + 16\pi^2 N^2 c^2} = c\sqrt{1 + 16\pi^2 N^2}$. Comparing, $k=16$.