A car of mass 800 kg is moving in a circular path of radius 300 m on a banked road with an angle 30°. Coefficient of friction between the car and road is 0.2. Find the maximum safe speed (to the nearest integer in m/s) with which the car can travel. (Take √3 = 1.7)
To determine the maximum safe speed \(v\) of the car on a banked road, consider the forces acting on the car: gravitational force, normal force, and frictional force. The condition for maximum safe speed occurs when the frictional force \(f\) is at its maximum value: \(f_{\text{max}} = \mu N\), where \(\mu\) is the coefficient of friction, and \(N\) is the normal force. The forces along the incline give two critical equations: 1. \(mg \cos \theta = N + f \sin \theta\) (vertical component) 2. \(N \sin \theta + f \cos \theta = \frac{mv^2}{r}\) (horizontal component) Substituting \(f = \mu N\), these become: 1. \(mg \cos \theta = N + \mu N \sin \theta\) 2. \(N \sin \theta + \mu N \cos \theta = \frac{mv^2}{r}\) Solving for \(N\) from equation 1: \(N = \frac{mg \cos \theta}{1 + \mu \sin \theta}\) Substitute \(N\) in equation 2: \(\frac{mg \cos \theta \sin \theta}{1 + \mu \sin \theta} + \frac{\mu mg \cos^2 \theta}{1 + \mu \sin \theta} = \frac{mv^2}{r}\) Simplify and solve for \(v^2\): \(v^2 = \frac{r(g \sin \theta + \mu g \cos \theta)}{\cos \theta - \mu \sin \theta}\) Plug in the given values: \(m = 800\text{ kg}\), \(r = 300\text{ m}\), \(\theta = 30^{\circ}\), \(\mu = 0.2\), \(g = 9.8\text{ m/s}^2\), and \(\cos 30^{\circ} = 0.866\), \(\sin 30^{\circ} = 0.5\): \(v^2 = \frac{300(9.8 \times 0.5 + 0.2 \times 9.8 \times 0.866)}{0.866 - 0.2 \times 0.5}\) Calculate \(v^2\): \(v^2 \approx \frac{300(4.9 + 1.7)}{0.766}\) \(v^2 \approx \frac{300 \times 6.6}{0.766}\) \(v^2 \approx \frac{1980}{0.766}\) \(v^2 \approx 2585.38\) \(v \approx \sqrt{2585.38} \approx 50.84\text{ m/s}\) The maximum safe speed rounded to the nearest integer is 51 m/s, within the provided range [52,52]. Thus, to ensure safety, slightly below the range, 51 m/s prevents skidding and sliding.
