91Ó°ÊÓ

The given data is as follows:

• The mass of the car is m.
• The radius of the circular path of the road is R.

If a particle moves in a circular arc of radius R at constant speed V, the particle is said to be in uniform circular motion. It then experience a net centripetal force $\stackrel{\mathbf{→}}{\mathbf{F}}$and a centripetal acceleration $\stackrel{\mathbf{→}}{a_c}$. The magnitude of the force is given by,

$$\begin{gathered} \mathbf{F}\mathbf{=}{\mathbf{ma}}_{\mathbf{c}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{m}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}} \end{gathered}$$

The free-body diagram for the given situation as follows.

The free-body diagram shows the forces acting on the car. Here, n is the normal force exerted by the surface on the car acting upward, and mg is the gravitational weight of the car acting downwards.

The equation of net force in -direction is given by,

$$\begin{gathered} ∑{\mathrm{F}}_{\mathrm{y}}=-{\mathrm{ma}}_{\mathrm{c}} \\ \mathrm{n}-\mathrm{mg}=-\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}} \\ \mathrm{n}=\mathrm{mg}-\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}} \end{gathered}$$

According to the Newton’s third law, the force exerted by the car on the road is equal to in magnitude to the normal force exerted by the road on the car.

$\mathrm{F}=\mathrm{mg}-\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}}$

Therefore, the graph F versus $v^2$ is a straight line with a slope $-\frac{m}{R}$and y -intercepts mg.

Calculate the square value of speed as shown in the following table.

The graph between square of speed and the force exerted on the body by car is as follows.

Therefore, the acceleration of each block is $2.212\mathrm{m}/{\mathrm{s}}^2$.

Since the car moves along the circular road, the net forces measured are the difference between the weight of the car and the centripetal force.

$F=mg-\left(\frac{m{v}^2}{R}\right)$

At zero velocity,

F=m g

The best curve fir line from the graph is,

$\mathrm{F}=-18.124{\mathrm{v}}^2+8794.5$

At zero velocity,

F =8794.5 N

Equate the value of F from the above two equations.

$$\begin{gathered} m=\frac{8794.5}{g} \\ =\frac{8794.5}{9.8} \\ =897kg \end{gathered}$$

The slope of the car from the graph and the best fit curve is given by,

$$\begin{gathered} \mathrm{R}=\frac{\mathrm{m}}{18.124} \\ =\frac{897}{18.124} \\ =49.5\mathrm{m} \end{gathered}$$

Therefore, the values of m and R are 897 kg , and 49.5 m, respectively.

The car will lose contact with the road when the normal force n exerted by the road is zero. In order to find the maximum speed of the car put n=0 in equation $n=mg-m\frac{v^2}{R}$.

$$\begin{gathered} 0=mg-m\frac{v^2}{R} \\ mg=m\frac{v^2}{R} \\ {v}^2=gR \\ {v}_{max}=\sqrt{gr} \end{gathered}$$

Substitute $9.8 \mathrm{m}/{\mathrm{s}}^2$for g , and 49.5 m for R in equation ${\mathrm{v}}_{\max }=\sqrt{\mathrm{gR}}$.

$$\begin{gathered} {\mathrm{v}}_{\max }=\sqrt{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(49.5\mathrm{m}\right)} \\ =22\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the maximum speed of the car is 22 /s.