Numerical Problems Based on Banking of Roads and Bending of a Cyclist for Class 11 Physics

Radius of curve, $r = 30\text{ m}$; Coefficient of friction, $\mu = 0.4$; $g = 10\text{ ms}^{-2}$.

The maximum safe speed on an unbanked level road is given by the formula:

Speed, $v = 60\text{ km h}^{-1} = 60 \times \frac{5}{18} = \frac{50}{3}\text{ ms}^{-1}$. Radius, $r = 40\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

For a safe turn on a level road, $v \le \sqrt{\mu rg}$. Thus, the minimum coefficient of friction required is:

We first calculate the maximum safe velocity ($v_{\max}$) allowed for this turn.

Radius, $r = 100\text{ m}$; $\mu = 0.6$; $g = 10\text{ ms}^{-2}$.

Since the rider’s actual speed ($10\text{ ms}^{-1}$) is much less than the maximum safe speed ($24.5\text{ ms}^{-1}$), he will easily be able to cross the turn without skidding.

Speed, $v = 14\sqrt{3}\text{ ms}^{-1}$; Radius, $r = 20\sqrt{3}\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

The angle of inclination ($\theta$) from the vertical for a cyclist is given by $\tan\theta = \frac{v^2}{rg}$.

Speed, $v = 6\text{ ms}^{-1}$; Radius, $r = 18\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

1. Calculate angle $\theta$:

2. Minimum coefficient of friction ($\mu$):
For the cyclist not to slip, the friction must at least match the required centripetal force component, which geometrically means $\mu \ge \tan\theta$.

Diameter $= 320\text{ m} \implies$ Radius, $r = 160\text{ m}$.

Speed, $v = 144\text{ km h}^{-1} = 144 \times \frac{5}{18} = 40\text{ ms}^{-1}$.

The angle of leaning ($\theta$) is given by:

Speed, $v = 500\text{ kmh}^{-1} = 500 \times \frac{5}{18} = \frac{1250}{9}\text{ ms}^{-1} \approx 138.89\text{ ms}^{-1}$.

Angle, $\theta = 30^{\circ}$. Take $g = 9.8\text{ ms}^{-2}$.

When an aircraft banks, the lift force provides the centripetal acceleration, obeying $\tan\theta = \frac{v^2}{rg}$. Rearranging for radius ($r$):

Speed, $v = 60\text{ kmh}^{-1} = 60 \times \frac{5}{18} = \frac{50}{3}\text{ ms}^{-1}$.

Radius, $r = 0.1\text{ km} = 100\text{ m}$.

The optimum angle of banking ($\theta$) is calculated ignoring friction:

Height of CG, $h = 1\text{ m}$; Distance between rails, $a = 1\text{ m}$; Radius, $r = 80\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

To avoid overturning on an unbanked curve, the overturning moment caused by the centrifugal force must not exceed the restoring moment of the carriage’s weight about the outer rail.

Radius, $r = 800\text{ m}$. Width of road, $w = 19.6\text{ m}$. Height difference, $h = 1\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

From the geometry of the banked road, $\sin\theta = \frac{h}{w} = \frac{1}{19.6}$. Since $h$ is very small compared to $w$, the angle $\theta$ is small, and we can approximate $\tan\theta \approx \sin\theta = \frac{1}{19.6}$.

Using the banking formula $v^2 = rg \tan\theta$:

Radius, $r = 400\text{ m}$. Speed, $v = 48\text{ kmh}^{-1} = 48 \times \frac{5}{18} = \frac{40}{3}\text{ ms}^{-1}$. Track width, $x = 1\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

First, find the required banking angle ($\theta$):

For small angles, the height raised ($h$) is given by $h = x \tan\theta$ (or $x \sin\theta$ for small angles):

Radius, $r = 750\text{ m}$; Bank angle, $\theta = 5^{\circ}$; Coefficient of friction, $\mu = 0.5$. Take $g = 9.8\text{ ms}^{-2}$.

Note: $\tan(5^{\circ}) \approx 0.0875$.

The maximum safe speed on a banked curve with friction is given by: