Question

A particle $P$ is moving on a circle under the action of only one force acting always towards fixed point $O$ on the circumference. Find ratio of $\frac{d^2\theta }{d{t}^2}$ and ${\left(\frac{d\theta }{dt}\right)}^2$.

Answer 1

Given,

A particle moving in a circle.

So, Let the radius of the circle is R,

So total length traveled by the object is $2\pi R$

Time taken$=T$

$v=\frac{d\theta }{dt}=\frac{2\pi r}{T}=f\times 2\pi R\Rightarrow (\frac{d\theta }{dt}{)}^2=(f\times 2\pi R{)}^2\Rightarrow \frac{d^2\theta }{d{t}^2}=a=\frac{v^2}{R}=\frac{(f\times 2\pi R{)}^2}{R}$

So the Ratio is $\frac{d^2\theta }{d{t}^2}to(\frac{d\theta }{dt}{)}^2=\frac{(f\times 2\pi R{)}^2}{(f\times 2\pi R{)}^2\times R}=\frac{1}{R}$