91Ó°ÊÓ

The passenger of mass m riding around a horizontal circle of radius$r$ at speed $v$.

The problem deals with Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body. Also, it deals with the centripetal force. It is a force that makes a body follow a curved path.

Formula:

Centripetal force is given by,

$\mathrm{F}={\mathrm{mv}}^2/\mathrm{r}$

The centripetal force on passengers is,

$\mathrm{F}={\mathrm{mv}}^2/\mathrm{r}$

The variation of F with respect to r while holding v constant is,

$\mathrm{dF}=−\frac{{\mathrm{mv}}^2}{{\mathrm{r}}^2}\mathrm{dr}$

Thus, variation in the net force$\text{d}\mathrm{F}$ for variation in radius $\text{d}\mathrm{r}$with $v$constant is $−\frac{{\mathrm{mv}}^2}{{\mathrm{r}}^2}\mathrm{dr}$.

The centripetal force on passengers is,

$\mathrm{F}={\mathrm{mv}}^2/\mathrm{r}$

The variation of F with respect to v while holding r constant is,

$\mathrm{dF}=\frac{2\mathrm{mv}}{\mathrm{r}}\mathrm{dv}$

Thus, variation in the net force $\text{d}\mathrm{F}$for variation in speed $\text{d}\mathrm{v}$with $r$constant is $\frac{2\mathrm{mv}}{\mathrm{r}}\mathrm{dv}$.

The period of the circular ride is,

$\mathrm{T}=2\mathrm{Ï€°ù}/\mathrm{v}$

Thus,

$\begin{array}{c}\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}\\ =\frac{\mathrm{m}}{\mathrm{r}}{\left(\frac{2\mathrm{Ï€°ù}}{\mathrm{v}}\right)}^2\\ =\frac{4{\mathrm{Ï€}}^2\mathrm{mr}}{{\mathrm{T}}^2}\end{array}$

Hence the variation of F with respect to T while holding r constant is,

$\begin{array}{l}\mathrm{dF}=−\frac{8{\mathrm{Ï€}}^2\mathrm{mr}}{{\mathrm{T}}^3}\mathrm{dT}\\ \mathrm{dF}=−8{\mathrm{Ï€}}^2\mathrm{mr}{\left(\frac{\mathrm{v}}{2\mathrm{Ï€°ù}}\right)}^3\mathrm{dT}\\ \mathrm{dF}=−\left(\frac{{\mathrm{mv}}^3}{{\mathrm{Ï€°ù}}^2}\right)\mathrm{dT}\end{array}$

Thus, variation in the net force$\text{d}\mathrm{F}$ for variation in period$\text{d}\mathrm{T}$with$r$constant is$−\left(\frac{{\mathrm{mv}}^3}{{\mathrm{Ï€°ù}}^2}\right)\mathrm{dT}$.