An${}^{\mathbf{85}\mathbf{.}\mathbf{0}\mathbf{kg}}$passenger is made to move along a circular path of radius${}^{\mathbf{r}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{m}}$in uniform circular motion. (a) Figure 6-40a is a plot of the required magnitude${}^{\mathbf{F}}$of the net centripetal force for a range of possible values of the passenger’s speed v. What is the plot’s slope at${}^{\mathbf{V}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$? (b) Figure 6-40b is a plot of F for a range of possible values of${}^{\mathbf{T}}$, the period of the motion. What is the plot’s slope atrole="math" localid="1654172716493" ${}^{\mathbf{T}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{s}}$?

Circular path of radius${}^{\mathrm{r}=3.50\mathrm{m}}$

Mass of passenger m= 85kg

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity. Also, it involves centripetal force which makes an object follow the circular path.

Formula:

The velocity in uniform circular motion is given by

$\mathrm{v}=\frac{2\mathrm{\pi r}}{\mathrm{T}}$ (i)

where, vis the velocity, r is the radius andTis the time period

the centripetal force is given by,

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$ (ii)

The slope of the plot at $\mathrm{v}=8.30\mathrm{m}/\mathrm{s}$is.

$$\begin{gathered} \frac{\mathrm{dF}}{\mathrm{dv}}=\frac{2\mathrm{mv}}{\mathrm{r}} \\ =\frac{2(85.0\mathrm{kg})(8.30\mathrm{m}/\mathrm{s})}{3.50\mathrm{m}} \\ =403\mathrm{N}.\mathrm{s}/\mathrm{m} \end{gathered}$$

Hence, the plot's slope is$403\mathrm{N}.\mathrm{s}/\mathrm{m}.$

The period of the circular ride is

$$\begin{gathered} \mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}} \\ \mathrm{Thus}, \\ \mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ =\frac{\mathrm{m}}{\mathrm{r}}{\left(\frac{2\mathrm{\pi r}}{\mathrm{T}}\right)}^2 \\ =\frac{4{\pi }^{2}mr}{T^2} \end{gathered}$$

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

$$\begin{gathered} \mathrm{dF}=-\frac{8{\mathrm{\pi }}^2\mathrm{mr}}{{\mathrm{T}}^3}\mathrm{dT} \\ \mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{plot}\mathrm{at}\mathrm{T}=2.50\mathrm{s}\mathrm{is}. \\ \frac{\mathrm{dF}}{\mathrm{dT}}=-\frac{8{\mathrm{\pi }}^2\mathrm{mr}}{{\mathrm{T}}^3} \\ =\frac{-8{\mathrm{\pi }}^2(85.0\mathrm{kg})(3.50\mathrm{m})}{{(2.50\mathrm{s})}^3} \\ =-1.50\times {10}^3\mathrm{N}/\mathrm{s} \\ \mathrm{Hence},\mathrm{the}\mathrm{period}\mathrm{of}\mathrm{the}\mathrm{motion}\mathrm{and}\mathrm{the}\mathrm{plot}\text{'}\mathrm{s}\mathrm{slope}\mathrm{at}\mathrm{T}=2.50\mathrm{s}\mathrm{is}\mathrm{T}=2\mathrm{\pi r}/\mathrm{v}\mathrm{and} \\ -1.50\times {10}^3\mathrm{N}/\mathrm{s} \end{gathered}$$