Question

homework 5 begin date: 8/17/2025 11:00:00 pm due date: 9/30/2025 11:59:00 pm end date: 9/30/2025 11:59:00 pm problem 9: (8% of assignment value) a car is driving along a level and unbanked circular track of diameter d at a constant speed of v. part (a) write an expression for the magnitude of the acceleration of the car in terms of the given parameters. a = 2v²/d ✓ correct! part (b) the diameter of the track and speed of the car are measured and it is found that d = 940 m and v = 29.7 m/s. what is the magnitude, in meters per squared second, of the acceleration of the car? a = 1.880 m/s² ✓ correct! part (c) write an expression for the minimum coefficient of friction between the cars tires and the road that is required in order to keep the car going in a circle in terms of the given parameters. part (d) what is the value of the minimum coefficient of friction between the cars tires and the road that is required in order to keep the car going in a circle? μmin = grade summary deductions potential 0% 100% submissions attempts remaining: 3 (4% deduction per attempt) detailed view

Explanation:

Step1: Recall centripetal - force formula

The centripetal acceleration \(a = \frac{v^{2}}{r}\), and given \(r=\frac{d}{2}\), so \(a = \frac{2v^{2}}{d}\) (which is already given in part (a)).

Step2: Calculate acceleration in part (b)

Given \(d = 940\ m\) and \(v = 29.7\ m/s\). Substitute into \(a=\frac{2v^{2}}{d}\). First, calculate \(v^{2}=(29.7)^{2}=882.09\). Then \(2v^{2}=2\times882.09 = 1764.18\). Now, \(a=\frac{1764.18}{940}\approx1.88\ m/s^{2}\).

Step3: Analyze centripetal - force and friction

The centripetal force \(F_c\) required for the car to move in a circle is provided by the frictional force \(F_f\). The centripetal force \(F_c=ma\) and the frictional force \(F_f=\mu_sN\), where \(N = mg\) (on a level road, normal force equals the weight). So \(ma=\mu_smg\), and \(\mu_s=\frac{a}{g}\).

Step4: Calculate the coefficient of friction

Given \(a = 1.88\ m/s^{2}\) and \(g = 9.8\ m/s^{2}\). Then \(\mu_s=\frac{1.88}{9.8}\approx0.192\).

Answer:

\(\mu_{min}=0.192\)