5.3 By both friction and banking of road

When the jeep runs with velocity $\left( {v = \sqrt {gR\tan \theta } } \right)$, no friction will acts between car tires and ground.

In the above equation $g$ and $\tan \theta $ are constants.

Therefore, $$v \propto R$$

When velocity $v$ increases, radius $R$ of the circular turn increases and vice versa.

When $v > \sqrt {gR\tan \theta } $, jeep will tends to move outward, so that the radius $R$ of the circular turn increases.

Due to this friction will act inwards to oppose the motion as shown in the figure.

From FBD we get, $$\begin{equation} \begin{aligned} N\sin \theta + f\cos \theta = \frac{{m{v^2}}}{R}\quad ...(i) \\ N\cos \theta = mg + f\sin \theta \quad ...(ii) \\\end{aligned} \end{equation} $$

When velocity $v$ of the jeep increases, friction between tires of jeep and ground increases.

Coefficient of friction between the jeep tires and road is $\mu $.

So, for safe turning of jeep at maximum velocity $v=v_max$, limiting friction will act between the tires and ground.

Therefore, $$f = {f_L} = \mu N\quad ...(iii)$$

From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} N\sin \theta + \mu N\cos \theta = \frac{{mv_{\max }^2}}{R} \\ N(\sin \theta + \mu \cos \theta ) = \frac{{mv_{\max }^2}}{R}\quad ...(iv) \\ N\cos \theta = mg + \mu N\sin \theta \\ N(\cos \theta - \mu \sin \theta ) = mg\quad ...(v) \\\end{aligned} \end{equation} $$

Dividing equation $(iv)$ and $(v)$ we get,

$$\begin{equation} \begin{aligned} \frac{{\left( {\sin \theta + \mu \cos \theta } \right)}}{{\left( {\cos \theta - \mu \sin \theta } \right)}} = \frac{{v_{\max }^2}}{{gR}} \\ {v_{\max }} = \sqrt {gR\frac{{\left( {\sin \theta + \mu \cos \theta } \right)}}{{\left( {\cos \theta - \mu \sin \theta } \right)}}} \\\end{aligned} \end{equation} $$

Dividing the both LHS and RHS by $\cos \theta $ we get,

$${v_{\max }} = \sqrt {gR\frac{{\left( {\tan \theta + \mu } \right)}}{{\left( {1 - \mu \tan \theta } \right)}}} $$

When $v < \sqrt {gR\tan \theta } $, jeep will tends to move inwards, so that the radius $R$ of the circular turn decreases.

Due to this friction will act outwards to oppose the motion as shown in the figure.

From FBD we get, $$\begin{equation} \begin{aligned} N\sin \theta - f\cos \theta = \frac{{m{v^2}}}{R}\quad ...(i) \\ N\cos \theta + f\sin \theta = mg\quad ...(ii) \\\end{aligned} \end{equation} $$

When velocity $v$ of the jeep decreases, friction between tires of jeep and ground increases.

Coefficient of friction between the jeep tires and road is $\mu $.

So, for safe turning of jeep at minimum velocity $v=v_min$, limiting friction will act between the tires and ground.

Therefore, $$f = {f_L} = \mu N\quad ...(iii)$$

From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} N\sin \theta - \mu N\cos \theta = \frac{{mv_{\min }^2}}{R} \\ N(\sin \theta - \mu \cos \theta ) = \frac{{mv_{\min }^2}}{R}\quad ...(iv) \\ N\cos \theta + \mu N\sin \theta = mg \\ N(\cos \theta + \mu \sin \theta ) = mg\quad ...(v) \\\end{aligned} \end{equation} $$

Dividing equation $(iv)$ and $(v)$ we get,

$$\begin{equation} \begin{aligned} \frac{{\left( {\sin \theta - \mu \cos \theta } \right)}}{{\left( {\cos \theta + \mu \sin \theta } \right)}} = \frac{{v_{\min }^2}}{{gR}} \\ {v_{\min }} = \sqrt {gR\frac{{\left( {\sin \theta - \mu \cos \theta } \right)}}{{\left( {\cos \theta + \mu \sin \theta } \right)}}} \\\end{aligned} \end{equation} $$

Dividing the both LHS and RHS by $\cos \theta $ we get,

$${v_{\min }} = \sqrt {gR\frac{{\left( {\tan \theta - \mu } \right)}}{{\left( {1 + \mu \tan \theta } \right)}}} $$

So, for safe turning on a road banked with angle $\theta $, the velocity of jeep should be, $$v \geqslant \sqrt {gR\frac{{\left( {\tan \theta - \mu } \right)}}{{\left( {1 + \mu \tan \theta } \right)}}} $$

Question 6. A jeep can take a safe turn on a road banked with angle $45^\circ $ with minimum and maximum speed have $36\ km/h$ and $108\ km/h$ respectively. Find the radius of the circular turn and coefficient of friction between the tires of jeep and ground.

Solution: Given, $$\begin{equation} \begin{aligned} {v_{\min }} = 36\;km/h = 10\;m/s \\ {v_{\max }} = 108\;km/h = 30\;m/s \\ \theta = 45^\circ \\\end{aligned} \end{equation} $$

As we know, $$\begin{equation} \begin{aligned} {v_{\min }} = \sqrt {gR\frac{{\left( {\tan \theta - \mu } \right)}}{{\left( {1 + \mu \tan \theta } \right)}}} \\ {v_{\max }} = \sqrt {gR\frac{{\left( {\tan \theta + \mu } \right)}}{{\left( {1 - \mu \tan \theta } \right)}}} \\\end{aligned} \end{equation} $$

We can write, $$\begin{equation} \begin{aligned} 10 = \sqrt {\frac{{gR(1 - \mu )}}{{(1 + \mu )}}} \quad ...(i) \\ 30 = \sqrt {\frac{{gR(1 + \mu )}}{{(1 - \mu )}}} \quad ...(ii) \\\end{aligned} \end{equation} $$

Dividing equation $(ii)$ by $(i)$ we get,

$$\begin{equation} \begin{aligned} 3 = \left( {\frac{{1 + \mu }}{{1 - \mu }}} \right) \\ \mu = 0.5\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} 10 = \sqrt {\frac{{10R(1 - 0.5)}}{{(1 + 0.5)}}} \\ 100 = \frac{{10R(0.5)}}{{1.5}} \\ R = 30m \\\end{aligned} \end{equation} $$

Question 7. A car moves with a constant tangential acceleration $a_t$ on a horizontal circular track of radius $R$. The coefficient of sliding friction between the tires of the car and road is $\mu $. What distance will the car travel without sliding, if the car starts from rest.

Initial velocity: $u=0$

Tangential acceleration: $a_t$

Tangential acceleration $a_t$ will accelerate the car.

Let the car traveled a total distance $s$ without sliding.

Therefore, $${v^2} = {u^2} + 2as$$ So, $${v^2} = 2{a_t}s$$

Centripetal acceleration due to velocity $v$, $$\frac{{{v^2}}}{R}\quad {\text{or}}\quad \frac{{2{a_t}s}}{R}$$

From the figure,

$m{a_t}$: Tangential force

$\frac{{m{v^2}}}{R}$: Centripetal force

${F_{net}}$: Resultant of tangential and centripetal force

$${{\vec F}_{net}} = - \frac{{m{v^2}}}{R}\hat i + m{a_t}\hat j$$ or $$\begin{equation} \begin{aligned} {\overrightarrow F _{net}} = - \frac{{ - 2m{a_t}s}}{R}\widehat i + m{a_t}\widehat j \\ \left| {{{\overrightarrow F }_{net}}} \right| = \sqrt {{{\left( {\frac{{2m{a_t}s}}{R}} \right)}^2} + {{\left( {m{a_t}} \right)}^2}} \\\end{aligned} \end{equation} $$

This ${\overrightarrow F _{net}}$ force will be provided by friction. So, when the car is at the verge of sliding, limiting friction will act between the car tires and road.

$$\begin{equation} \begin{aligned} N = mg \\ {f_L} = \mu N \\ {f_L} = \mu mg \\\end{aligned} \end{equation} $$

Therefore, $$\begin{equation} \begin{aligned} \sqrt {{{\left( {\frac{{2m{a_t}s}}{R}} \right)}^2} + {{\left( {m{a_t}} \right)}^2}} = \mu mg \\ s = \frac{R}{2}\sqrt {{{\left( {\frac{{\mu g}}{{{a_t}}}} \right)}^2} - 1} \\\end{aligned} \end{equation} $$