Cheetahs running at top speed have been reported at an astounding$\text{114\hspace{0.17em}km/h}$(about$\text{71\hspace{0.17em}mi/h}$) by observers driving alongside the animals. Imagine trying to measure a cheetah’s speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering$\text{114\hspace{0.17em}km/h}$. You keep the vehicle a constantfrom the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius$\text{92\hspace{0.17em}m}$. Thus, you travel along a circular path of radius.$\text{100\hspace{0.17em}m}$ (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah’s speed is$\text{114\hspace{0.17em}km/h}$, and that type of error was apparently made in the published reports.)

The rate of change of angular displacement with respect to time, is known as angular velocity. From velocity and radius of the path, find the angular speed of an observer and Cheetah around the circular paths using the corresponding formula for circular motion. From this, find the linear speed by using its relation with angular speed.

The angular velocity is given as-

$\mathbf{\omega }\mathbf{=}\frac{v}{R}$

where, v is velocity, R is radius and $\mathbf{\omega }$is angular frequency.

Speed of the observer is,
$\begin{array}{c}v=114\frac{\text{km}}{\text{h}}\\ =114(\frac{5}{18}\text{m}/\text{s})\\ =31.67\text{m}/\text{s}.\end{array}$

Since, an observer is abreast of the Cheetah, bothhave the sameangular speed.

The angular speed of an observer and Cheetah around the circular paths is,

$\omega =\frac{v}{r}$

$\omega =\frac{31.67\text{m/s}}{100\text{\hspace{0.17em}m}}$

$\omega =0.32\text{\hspace{0.17em}rad/s}$

Hence, the angular speed of an observer and Cheetah around the circular paths is.$\omega =0.32\text{\hspace{0.17em}rad/s}$

Linear speed of Cheetah along its path is,

$\begin{array}{c}v=R\omega \\ =92\text{\hspace{0.17em}m}(0.32\text{\hspace{0.17em}rad}/\text{s})(\frac{18}{5}\text{km/h})\\ =1.05\times {10}^2\text{km/h}\end{array}$

Hence, the linear speed of Cheetah along its paths is$1.05\times {10}^2\text{km}/\text{h}.$