Q.A circular racetrack of radius $300\ \text{m}$ 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

The key idea here is the Static Friction Limit, which dictates the maximum possible frictional force available to prevent slipping.

(a) For the optimum speed, no frictional force is required. The horizontal component of the normal force provides the necessary centripetal force.

1. Resolve the normal force $N$ into vertical and horizontal components.
2. Vertical equilibrium: $N \cos \theta = mg$.
3. Horizontal centripetal force: $N \sin \theta = \frac{mv^2}{r}$.
4. Dividing these equations gives the optimum speed $v = \sqrt{rg \tan \theta}$. Substituting $r=300\ \text{m}$, $g=9.8\ \text{m/s}^2$, $\theta=15^{\circ}$: $v = \sqrt{300 \times 9.8 \times \tan 15^{\circ}} \approx \sqrt{2940 \times 0.2679} \approx \sqrt{787.626} \approx 28.06\ \text{m/s}$.

(b) For the maximum permissible speed, the car tends to slip outwards (up the incline), so the static frictional force $f_s = \mu_s N$ acts inwards (down the incline), assisting the normal force in providing the centripetal force.

1. Resolve all forces (weight $mg$, normal force $N$, and friction $f_s$) into vertical and horizontal components.
2. Vertical equilibrium: $N \cos \theta - f_s \sin \theta = mg$.
3. Horizontal centripetal force: $N \sin \theta + f_s \cos \theta = \frac{mv_{max}^2}{r}$.
4. Substitute $f_s = \mu_s N$ and solve for $v_{max}$: $v_{max} = \sqrt{rg \left( \frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} \right)}$. …

When a vehicle navigates a curve, it requires a centripetal force to change its direction. On a flat road, this force is entirely provided by friction. However, on a banked road, the road surface is tilted, allowing a component of the normal force to contribute to the centripetal force. This reduces the reliance on friction, making turns safer and more comfortable, especially at higher speeds.

The problem asks for two distinct speeds:

1. Optimum Speed: This is the specific speed at which the horizontal component of the normal force exactly provides the necessary centripetal force. At this speed, no frictional force is required, which means there is no wear and tear on the tyres.
2. Maximum Permissible Speed: This is the highest speed a car can achieve without slipping. At this speed, the car is on the verge of slipping up the bank, so the maximum static friction acts down the incline, assisting the normal force in providing the centripetal force.

Let's analyze the forces acting on the car in both scenarios. We will use a coordinate system where the x-axis is horizontal (towards the center of the track) and the y-axis is vertical.

Given Data:

• Radius of the track, $R = 300\ \text{m}$
• Banking angle, $\theta = 15^{\circ}$
• Coefficient of static friction, $\mu_s = 0.2$
• Acceleration due to gravity, $g = 9.8\ \text{m/s}^2$

(a) Optimum Speed of the Race-car

At the optimum speed, the car does not rely on friction. The centripetal force is provided entirely by the horizontal component of the normal force.

1. Identify Forces and Resolve Components:

   • Gravitational Force ($mg$): Acts vertically downwards.
   • Normal Force ($N$): Acts perpendicular to the banked surface.
     • Its vertical component is $N \cos\theta$, acting upwards.
     • Its horizontal component is $N \sin\theta$, acting towards the center of the track.

2. Apply Newton's Second Law:

   • Vertical Equilibrium: Since the car is not accelerating vertically, the net vertical force is zero.

*   **Horizontal Motion:** The net horizontal force provides the centripetal force ($F_c = \frac{mv^2}{R}$).

3. Solve for $v_{opt}$: Divide equation (2) by equation (1):

Rearranging for $v_{opt}$:

> [!FORMULA]
> The optimum speed for a banked curve is given by:
> $$ v_{opt} = \sqrt{gR \tan\theta} $$

4. Substitute Values:

We know $\tan(15^{\circ}) \approx 0.2679$.

(b) Maximum Permissible Speed to Avoid Slipping

At the maximum permissible speed, the car is on the verge of slipping up the bank. Therefore, the maximum static friction force ($f_s = \mu_s N$) acts down the incline, helping to provide the necessary centripetal force.

1. Identify Forces and Resolve Components:

   • Gravitational Force ($mg$): Acts vertically downwards.
   • Normal Force ($N$): Acts perpendicular to the banked surface.
     • Vertical component: $N \cos\theta$ (upwards).
     • Horizontal component: $N \sin\theta$ (towards the center).
   • Static Friction Force ($f_s$): Acts parallel to the banked surface, down the incline.
     • Vertical component: $f_s \sin\theta$ (downwards).
     • Horizontal component: $f_s \cos\theta$ (towards the center).

2. Apply Newton's Second Law:

   • Vertical Equilibrium: The net vertical force is zero.

*   **Horizontal Motion:** The net horizontal force provides the centripetal force. Both the normal force and friction contribute.

3. Substitute Maximum Static Friction: Since the car is at the verge of slipping, $f_s = \mu_s N$. Substitute this into equations (3) and (4):
   • From (3): $N \cos\theta = mg + \mu_s N \sin\theta$ …

Concept: Banked Circular Track — with and without relying on friction

Step 1 (part a): Vertical and horizontal equilibrium at the optimum speed (no friction needed)

Step 2: Divide (2) by (1) to eliminate $N$

Step 3: Substitute the numbers

Step 4 (part b): At the maximum permissible speed, friction acts down the incline at its limiting value $f_s=\mu_s N$

Vertical: $N\cos\theta = mg + \mu_s N\sin\theta$. Horizontal: $N\sin\theta + \mu_s N\cos\theta = \dfrac{mv_{max}^2}{R}$. …