Circular Motion and Banking of Roads

Dynamics of Uniform Circular Motion

In circular motion, an object moves along a curved path with a constant or varying speed.For an object to be in uniform circular motion (moving at a constant speed v in a circle of radius r), it must have a net force acting on it that is directed towards the center of the circle. This net force is called the centripetal force ($F_c$).

From Newton's Second Law, the magnitude of this force is:

$$F_c = ma_c = \frac{mv^2}{r}$$

Where:

• $m$ is mass, $v$ is speed, $r$ is radius of the circular path

It is crucial to understand that centripetal force is not a new kind of force. It is the net result of other familiar forces like tension (in a string), gravity (for satellites), normal force (in banking), or friction (for a car on a flat road).

Characteristics:

• Always directed towards the center of the circle
• Provided by tension, friction, gravity, or normal force depending on the situation

Banking of Roads

Banking is the process of raising the outer edge of a curved road to provide necessary centripetal force for turning vehicles. This prevent cars from skidding on circular turns. The horizontal component of the normal force provides some or all of the necessary centripetal force. The ideal speed (or 'safe speed') for a banked curve (without friction) is the speed at which the horizontal component of the normal force exactly provides the centripetal force needed. $$N\sin\theta = \frac{mv^2}{r}$$ $$N\cos\theta = mg$$ Dividing these gives $$\tan\theta = \frac{v^2}{rg}$$

Banking Without Friction: $$\tan\theta = \frac{v^2}{rg}$$ Where $\theta$ is banking angle, $v$ is speed, $r$ is radius, $g$ is gravity.

Banking With Friction: $$\tan\theta = \frac{v^2}{rg} + \mu$$ (Approximate, exact derivation involves trigonometric resolution)

Applications:

• Design of curved roads, flyovers, race tracks
• Ensures safety at turns and minimizes dependence on friction

Example:

Question: A car of mass 1000 kg is moving at 18 m/s on a curve of radius 30 m. What is the centripetal force required?

Solution: $$F_c = \frac{mv^2}{r} = \frac{1000 \times (18)^2}{30} = \frac{324000}{30} = 10800\, N$$

Example (Banking of Roads):

A circular racetrack of radius 300 m is banked at an angle of $15^\circ$. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the maximum permissible speed to avoid slipping?

Solution: The forces on the car are gravity (mg), normal force (N), and friction (f). For maximum speed, the car tends to slip up the incline, so the frictional force acts down the incline.

1. Resolve forces vertically and horizontally:

   Vertical equilibrium (no vertical acceleration): $N\cos\theta = mg + f\sin\theta$.

   Horizontal motion (centripetal force): $N\sin\theta + f\cos\theta = \frac{mv^2}{R}$.

2. Use the condition for maximum friction:

   For maximum speed, friction is at its limit: $f = f_{s,max} = \mu_s N$.

3. Substitute and solve for N from the vertical equation:

   $N\cos\theta - \mu_s N\sin\theta = mg \implies N = \frac{mg}{\cos\theta - \mu_s\sin\theta}$.

4. Substitute N and f into the horizontal equation:

   $N(\sin\theta + \mu_s\cos\theta) = \frac{mv_{max}^2}{R}$.

   $$\frac{mg(\sin\theta + \mu_s\cos\theta)}{\cos\theta - \mu_s\sin\theta} = \frac{mv_{max}^2}{R}$$

   Dividing the numerator and denominator of the fraction by $\cos\theta$ gives the standard formula:

   $$v_{max} = \sqrt{Rg \frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}}$$

5. Plug in the values:

   $R=300, g=9.8, \theta=15^\circ, \mu_s=0.2$. $\tan(15^\circ) \approx 0.268$. $$v_{max} = \sqrt{300 \times 9.8 \frac{0.268 + 0.2}{1 - 0.2 \times 0.268}} = \sqrt{2940 \frac{0.468}{0.9464}} \approx \sqrt{1451.6} \approx 38.1\ m/s$$