Question

An automobile moves on a level horizontal road in a circle of radius 40.0 m. The coefficient of friction between tires and road is 0.60. The maximum speed with which this car can round this curve is

Answer 1

Approximately
`15\; {\rm m\cdot s^(-1)}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Let
`m` denote the mass of this vehicle, and let
`g` denote the gravitational field strength.

Since the road is level, the normal force that the road exerts on the vehicle would be equal in magnitude to the weight
`m\, g` of this vehicle. That is:
`F_\text{normal} = m\, g`.

Let
`\mu_(k)` denote the coefficient of (kinetic) friction between the tires of this vehicle and the road. The maximum friction that the road could exert on this vehicle would be:

`\begin{aligned}& F_\text{max. friction} && (\text{max. friction}) \\ =\; &\mu_(k)\, F_\text{normal} && (\text{coefficient of friction}) \,(\text{normal force}) \\ =\; & \mu_(k)\, m\, g\, \end{aligned}`.

Also because this road is level, the only unbalanced force on this vehicle would be friction. Thus, the net force on this vehicle would be equal to the friction that the road exerts on the vehicle, which is at most
`\mu_(k)\, m\, g`:

`\begin{aligned}& F_\text{net} && (\text{net force}) \\ =\; & F_\text{friction} && (\text{friction}) \\ \le \; & F_\text{max. friction} && (\text{max. friction})&\\ =\; & \mu_(k)\, m\, g\, \end{aligned}`.

In other words,
`F_\text{net} < \mu_(k)\, m\, g`. This inequality could provide an upper bound on the acceleration of this vehicle:

`\begin{aligned}& a && (\text{acceleration}) \\ =\; & \frac{F_\text{net}}{m} && \genfrac{}{}{0em}{}{(\text{net force})}{(\text{mass})}\\ \le \; & (\mu_(k)\, m\, g)/(m) \\ =\; & \mu_(k)\, g\end{aligned}`.

Thus, the maximum acceleration of this vehicle on this road would be
`a_\text{max} = \mu_(k)\, g`.

Let
`r` denote the radius of this circle. When the vehicle is moving around this circle at a speed of
`v`, the (centripetal) acceleration of this vehicle would continuously be
`a = (v^(2) / r)`.

Since the upper bound of the acceleration of this vehicle is
`a_\text{max} = \mu_(k)\, g`, rearrange the equation
`a = (v^(2) / r)` to find an upper bound on speed
`v`:

`\begin{aligned}(v^(2))/(r) &= a \\ & \le a_{\text{max}} \\ &= \mu_(k)\, g\end{aligned}`.

`v^(2) \le \mu_(k)\, g\, r`.

`v \le \sqrt{\mu_(k)\, g\, r}`.

In other words, when the radius of the circle is
`r`, the coefficient of friction between the tire and the ground is
`\mu_(k)`, and the gravitational field strength is
`g`, the maximum speed of the vehicle would be
`\sqrt{\mu_(k)\, g \, r}`.

In this question, it is given that
`r = 40.0\; {\rm m}` whereas
`\mu_(k) = 0.60`. Assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`, the maximum speed of this vehicle would be:

`\begin{aligned}v_\text{max} &= \sqrt{\mu_(k)\, g\, r} \\ &= \sqrt{0.60 * 9.81\; {\rm m\cdot s^(-2)} * 40.0\; {\rm m}} \\ &\approx 15\; {\rm m\cdot s^(-1)}\end{aligned}`.