NLM O37

Physics Solution Q37

Physics Solution: Banked Turn Analysis

1. Defining the Velocities

We are analyzing a car on a banked turn of radius $R$ and banking angle $\theta$ with coefficient of friction $\mu$.

  • Optimum Speed ($v_0$): The speed at which no friction is required. This occurs when the banking component of the normal force perfectly provides the centripetal force. $$ v_0 = \sqrt{Rg \tan \theta} $$
  • Maximum Speed ($v_{\text{max}}$): Friction acts down the slope to prevent sliding up. $$ v_{\text{max}} = \sqrt{Rg \left( \frac{\tan \theta + \mu}{1 – \mu \tan \theta} \right)} $$ Using the identity $\mu = \tan \phi$ (where $\phi$ is the angle of friction), this simplifies to: $$ v_{\text{max}} = \sqrt{Rg \tan(\theta + \phi)} $$
  • Minimum Speed ($v_{\text{min}}$): Friction acts up the slope to prevent sliding down. $$ v_{\text{min}} = \sqrt{Rg \left( \frac{\tan \theta – \mu}{1 + \mu \tan \theta} \right)} $$ Similarly simplifying: $$ v_{\text{min}} = \sqrt{Rg \tan(\theta – \phi)} $$
angle (α) v θ – φ v_min θ v_0 θ + φ v_max Avg

2. Comparing $v_0$ with the Average

We need to compare $v_0$ with the arithmetic mean of the maximum and minimum speeds: $$ \text{Average} = \frac{v_{\text{max}} + v_{\text{min}}}{2} $$

Let us consider the function $f(\alpha) = \sqrt{Rg \tan \alpha}$ assume the angle between 15 and 75 degrees. Then: $$ v_0 = f(\theta) $$ $$ v_{\text{max}} = f(\theta + \phi) $$ $$ v_{\text{min}} = f(\theta – \phi) $$

The arithmetic mean $\frac{f(\theta + \phi) + f(\theta – \phi)}{2}$ represents the midpoint of the secant line connecting two points on the curve. If the graph of $f(\alpha)$ is convex (concave up), the midpoint of the secant line will lie above the curve value at the midpoint $f(\theta)$.

Numerical Verification (e.g., $\theta = 45^\circ, \mu = 0.5$):
$v_0 = \sqrt{gR \cdot 1} = 1.0 \sqrt{gR}$
$v_{\text{max}} = \sqrt{gR \frac{1+0.5}{1-0.5}} = \sqrt{3gR} \approx 1.732 \sqrt{gR}$
$v_{\text{min}} = \sqrt{gR \frac{1-0.5}{1+0.5}} = \sqrt{gR/3} \approx 0.577 \sqrt{gR}$

Average $= \frac{1.732 + 0.577}{2} = 1.154 \sqrt{gR}$
Since $1.154 > 1.0$, we have: $$ \frac{v_{\text{max}} + v_{\text{min}}}{2} > v_0 $$

Conclusion: $v_0 < 0.5(v_{\text{max}} + v_{\text{min}})$

Correct Option: (a)