Let's solve the problem step by step as requested. --- # **A. Sketch the Situation & List Quantities** **Situation:** A car of mass \( m \) moves at constant speed \( v \) around a banked curve of radius \( R \), bank angle \( \theta \), with static friction coefficient \( \mu_s \) between tires and road. Gravity is \( g \). **Given/known:** - Mass: \( m \) - Speed: \( v \) - Curve radius: \( R \) - Banking angle: \( \theta \) - Static friction coefficient: \( \mu_s \) - Gravity: \( g \) **Coordinate System:** Let the \( x \)-axis point horizontally toward the center of the curve, and the \( y \)-axis perpendicular to the road surface. **Assumptions:** - The car is a point mass. - The speed is constant (circular motion). - Only static friction acts (no air resistance). **To Find:** - Maximum speed \( v_{max} \) (before sliding up the bank) - Minimum speed \( v_{min} \) (before sliding down the bank) --- # **B. Free-Body Diagram and Maximum Speed \( v_{max} \)** **Forces acting on the car:** - **Gravity:** \( mg \) (downward) - **Normal Force:** \( N \) (perpendicular to the road surface) - **Friction:** \( f \) (along the surface, direction depends on whether car slides up or down) **At maximum speed:** Friction acts **down** the incline (trying to prevent the car from sliding up the bank). **Free-Body Components:** - Perpendicular to the incline (vertical): \( N \cos\theta - f \sin\theta = mg \) - Parallel to the incline (horizontal, toward center): \( N \sin\theta + f \cos\theta = \frac{mv^2}{R} \) At the limit, \( f = \mu_s N \): **Plug in \( f = \mu_s N \) (down the bank for \( v_{max} \)):** \[ \begin{align*} N \cos\theta - \mu_s N \sin\theta &= mg \hspace{1cm} (1)\\ N \sin\theta + \mu_s N \cos\theta &= \frac{mv_{max}^2}{R} \hspace{1cm} (2) \end{align*} \] **Solve for \( N \) from (1):** \[ N (\cos\theta - \mu_s \sin\theta) = mg \implies N = \frac{mg}{\cos\theta - \mu_s \sin\theta} \] **Substitute \( N \) into (2):** \[ (\sin\theta + \mu_s \cos\theta)N = \frac{mv_{max}^2}{R} \] \[ (\sin\theta + \mu_s \cos\theta)\frac{mg}{\cos\theta - \mu_s \sin\theta} = \frac{mv_{max}^2}{R} \] **Solve for \( v_{max} \):** \[ v_{max}^2 = \frac{gR (\sin\theta + \mu_s \cos\theta)}{\cos\theta - \mu_s \sin\theta} \] \[ \boxed{ v_{max} = \sqrt{ \frac{gR (\sin\theta + \mu_s \cos\theta)}{\cos\theta - \mu_s \sin\theta} } } \] --- # **C. Minimum Speed \( v_{min} \)** **At minimum speed:** Friction acts **up** the incline (preventing the car from sliding down the bank). **Change sign of friction in equations:** - Perpendicular: \( N \cos\theta + \mu_s N \sin\theta = mg \) - Parallel: \( N \sin\theta - \mu_s N \cos\theta = \frac{mv_{min}^2}{R} \) Solve for \( N \): \[ N (\cos\theta + \mu_s \sin\theta) = mg \implies N = \frac{mg}{\cos\theta + \mu_s \sin\theta} \] Plug into parallel equation: \[ (\sin\theta - \mu_s \cos\theta)N = \frac{mv_{min}^2}{R} \] \[ (\sin\theta - \mu_s \cos\theta)\frac{mg}{\cos\theta + \mu_s \sin\theta} = \frac{mv_{min}^2}{R} \] Solve for \( v_{min} \): \[ v_{min}^2 = \frac{gR (\sin\theta - \mu_s \cos\theta)}{\cos\theta + \mu_s \sin\theta} \] \[ \boxed{ v_{min} = \sqrt{ \frac{gR (\sin\theta - \mu_s \cos\theta)}{\cos\theta + \mu_s \sin\theta} } } \] --- # **D. Sense-Making** ## **a. Check units:** - \( g \) has units of \(\text{m/s}^2\), \( R \) is in meters, trig functions are dimensionless. - So both \( v_{min}, v_{max} \) have units of \(\sqrt{\text{m/s}^2 \cdot \text{m}} = \text{m/s} \), which is correct. ## **b. Limiting case: frictionless surface (\( \mu_s = \))** **Plug in \( \mu_s = \) into both formulas:** - \( v_{max} = \sqrt{ \frac{gR (\sin\theta)}{\cos\theta} } = \sqrt{gR \tan\theta} \) - \( v_{min} = \sqrt{ \frac{gR (\sin\theta)}{\cos\theta} } = \sqrt{gR \tan\theta} \) Both formulas yield the same value, the **design speed** for a frictionless banked turn. --- # **E. Daytona International Speedway Application** ## **Given:** - \( R = 300\,\text{m} \) - \( \theta = 30^\circ \) - \( \mu_s = .8 \) - \( g = 9.8\,\text{m/s}^2 \) ### **a. Top speed (dry day, max speed):** Use the \( v_{max} \) formula: - \( \sin 30^\circ = .5 \), \( \cos 30^\circ = \sqrt{3}/2 \approx .866 \) \[ v_{max} = \sqrt{ \frac{9.8 \times 300 \times (.5 + .8 \times .866)}{.866 - .8 \times .5} } \] \[ .8 \times .866 = .6928 \] \[ \text{Numerator: } 9.8 \times 300 \times (.5 + .6928) = 9.8 \times 300 \times 1.1928 = 9.8 \times 357.84 = 3,509.832 \] \[ \text{Denominator: } .866 - .4 = .466 \] So: \[ v_{max} = \sqrt{ \frac{3,509.832}{.466} } = \sqrt{7,537.1} \approx 86.8\, \text{m/s} \] ### **b. Stationary car sliding down (call off race):** Stationary means \( v = \). Set the friction force up the incline to balance gravity **down** the incline: - The car will slide if gravity's component down the slope exceeds max friction. Down the incline: \( mg \sin\theta \) Up the incline: \( \mu_s N \) Normal force: \( N = mg \cos\theta \) So, maximum static friction: \( \mu_s mg \cos\theta \) Set the balance: \[ mg \sin\theta = \mu_s mg \cos\theta \implies \tan\theta = \mu_s \implies \mu_s = \tan 30^\circ = .577 \] If \( \mu_s < .577 \), the stationary car will slide down. --- # **Final Answers** --- ### **General Symbolic Answers** - **Maximum speed:** \[ \boxed{ v_{max} = \sqrt{ \frac{gR (\sin\theta + \mu_s \cos\theta)}{\cos\theta - \mu_s \sin\theta} } } \] - **Minimum speed:** \[ \boxed{ v_{min} = \sqrt{ \frac{gR (\sin\theta - \mu_s \cos\theta)}{\cos\theta + \mu_s \sin\theta} } } \] --- ### **Daytona Speedway** - **(a) Maximum speed with \( \mu_s = .8, \theta = 30^\circ, R = 300\,\text{m} \):** \[ \boxed{ v_{max} \approx 86.8 \,\text{m/s} } \] - **(b) Minimum friction for stationary car not to slide:** \[ \boxed{ \mu_s = \tan 30^\circ = .577 } \] If \( \mu_s < .577 \), call off the race. --- Let me know if you want a labeled diagram or further explanation!