Question

A point moves along a circle with a velocity v=kt, where $k=0.5m/s^2.$ Find the total acceleration of the point at the moment when it has covered the $n^{th}$ fraction of the circle after the beginning of motion, where $n=\frac{1}{10}.$

Answer 1

$v=\frac{ds}{dt}=kt$
or ${\int }_0^{s}ds=k{\int }_0^{t}tdt$
$\therefore s=\frac{1}{2}k{t}^2$
For completion of nth fraction of circle,
$s=2\pi rn$
or $t^2=\left(4\pi nr\right)/k$ ...(i)
Tangential acceleration $a_T=\frac{dv}{dt}=k$ ...(ii)
Normal acceleration $=a_N=\frac{v^2}{r}=\frac{k^2{t}^2}{r}$ ...(iii)
or $a_N=4\pi nk$
$\therefore a=\sqrt{(a_T^2+a_N^2)}={[k^2+16{\pi }^2{n}^2{k}^2]}^{1/2}$
$=k{[1+16{\pi }^2{n}^2]}^{1/2}$
$=0.50{[1+16\times {\left(3.14\right)}^2\times {\left(0.10\right)}^2]}^{1/2}$
$=0.8m/s^2$