91Ó°ÊÓ

• The radius of the circular curve, R = 200 m.
• The bank angle is$\mathrm{θ}$.
• The coefficient of friction is ${\mathrm{μ}}_s$.

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is related to the net force acting upon the object and the mass of the object.

Use Newton’s second law of motion with zero acceleration to find the expression of$V_{max}$in terms of bank angle and frictional coefficient.

Formula:

F = ma

The angle of$\stackrel{→}{F}$(measured from the vertical axis):

$\mathrm{ϕ}=\mathrm{θ}+{\mathrm{θ}}_s$

Where,

$\tan {\mathrm{θ}}_s={\mathrm{μ}}_s,\mathrm{and},\mathrm{θ}$is the bank angle.

The vector sum of $\stackrel{→}{F}$ and the vertically downward pull (mg) of gravity must be equal to the (horizontal) centripetal force $(m{v}^2/R)$:

$$\begin{gathered} \tan \mathrm{Ï•}=\frac{m{v}^2/R}{mg} \\ =\frac{v^2}{Rg} \end{gathered}$$

Substitute the expression of $\mathrm{Ï•}$in the above expression and simplify:

$$\begin{gathered} V_{max}^2=Rg\tan \left(\mathrm{θ}+{\tan }^{-1}{\mathrm{μ}}_s\right) \\ {V}_{max}=\sqrt{Rg×\frac{\tan \left(\mathrm{θ}\right)+{\mathrm{μ}}_s}{1-{\mathrm{μ}}_s\tan \left(\mathrm{θ}\right)}} \end{gathered}$$

Thus, the expression for v_max is $V_{max}=\sqrt{Rg×\frac{\tan \left(\mathrm{θ}\right)+{\mathrm{μ}}_s}{1-{\mathrm{μ}}_s\tan \left(\mathrm{θ}\right)}}$.

The graph is shown below (with $\mathrm{θ}$in radian):

Thus the graph is

For${\mathrm{μ}}_s=0.60$( the upper curve ), when$\mathrm{θ}=10°=0.175$radian, the max velocity can be calculated as:

$$\begin{gathered} V_{max}=\sqrt{200×9.8×\frac{\tan (10°)+0.60}{1-0.60×\tan (10°)}} \\ {V}_{max}=41.3\mathrm{m}/\mathrm{s} \\ {V}_{max}=149\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the value of max velocity is 149km/h.

For${\mathrm{μ}}_s=0.050$( the lower curve ), when$\mathrm{θ}=10°=0.175$radian, the max velocity can be calculated as:

$$\begin{gathered} V_{max}=\sqrt{200×9.8×\frac{\tan (10°)+0.50}{1-0.50×\tan (10°)}} \\ {V}_{max}=21.2\mathrm{m}/\mathrm{s} \\ {V}_{max}=76.2\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the value of max velocity is 76,2 km/h.