A police officer in hot pursuit drives her car through a circular turn of radius ${}^{\mathbf{300}\mathbf{}\mathbf{m}}$with a constant speed of ${}^{\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$. Her mass is${}^{\mathbf{55}\mathbf{.}\mathbf{0}\mathbf{kg}}$. What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces)

$\mathrm{Radius}=300\mathrm{m},\mathrm{speed}=80.\mathrm{km}/\mathrm{h},\mathrm{mass}=55.0\mathrm{kg}.$

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 Newton’s law of motion. Considering both horizontal and vertical component of the force, the magnitude of the net force can be calculated.

Formulae:

According to Newton’s law of motion the force can be written as

$F=\sum _{}ma$ (i)

where, F is the force, m is mass and a is an acceleration

Note that the speed 80.0 km/h in SI units is roughly 22.2 m/s. In this case, the horizontal force is equal to the centripetal force. The centripetal acceleration in this problem can be written as,

$a_c=\frac{v^2}{R}$

Using equation (i) the centripetal force can be written as,

$F_c=\frac{m{v}^2}{R}$ (ii)

Similarly, the upward force on her must be,

$F_{up}=mg$ (iii)

Adding equation (ii) and (iii) the magnitude of the net force can be written as,

role="math" localid="1654071502635" $$\begin{gathered} F_{net}=\sqrt[]{{F^2}_{up}+{F^2}_c} \\ =\sqrt[]{{\left(mg\right)}^2+{\left(m{v}^2/R\right)}^2} \\ =\sqrt[]{{\left(55.0\times \left(-9.8\right)\right)}^2+{\left(55.0\times 22.0^2/300\right)}^2} \\ =547\mathrm{N} \end{gathered}$$

Hence, the magnitude of the netforce of the officer on the car seat is 547 N.

As measured from the vertical axis,the angle is,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{F_{up}}{F_c}\right) \\ ={\tan }^{-1}\left(\frac{\left(m{v}^2/R\right)}{mg}\right) \\ ={\tan }^{-1}\left(\frac{v^2}{gR}\right) \\ ={\tan }^{-1}\left(\frac{22.0^2}{9.8\times 300}\right) \\ =9.{35}^{°} \end{gathered}$$

Hence, the angle (relative to vertical) of the netforce of the officer on the car seat is