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. μ_min = 1 grade summary deductions potential 0% 100% submissions attempts remaining: 3 4% deduction per attempt

Explanation:

Step1: Recall centripetal - acceleration formula

The centripetal acceleration formula is $a = \frac{v^{2}}{r}$, where $a$ is the centripetal acceleration, $v$ is the speed of the object, and $r$ is the radius of the circular path. Given that the diameter is $d$, then $r=\frac{d}{2}$. So, $a=\frac{2v^{2}}{d}$.

Step2: Calculate centripetal - acceleration for part (b)

We are given $d = 940$ m and $v = 29.7$ m/s. Substitute these values into the formula $a=\frac{2v^{2}}{d}$.
$a=\frac{2\times(29.7)^{2}}{940}=\frac{2\times882.09}{940}=\frac{1764.18}{940}\approx1.88$ m/s².

Step3: Find the minimum - coefficient of friction formula

The frictional force $F_f=\mu N$, where $N = mg$ (on a level road, normal force equals the weight of the car). The centripetal force is provided by the frictional force, so $F_c = F_f$. Also, $F_c=ma$ and $F_f=\mu N=\mu mg$. Then $ma=\mu mg$. Canceling out the mass $m$ of the car, we get $\mu=\frac{a}{g}$.

Answer:

a) $a = \frac{2v^{2}}{d}$
b) $a\approx1.88$ m/s²
c) $\mu=\frac{a}{g}$