Question

Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. B) Find the minimum static friction coefficient between the tires and the road.

Answer 1

At least
`0.128` (rounded up, to three significant figures as in speed,) assuming that the road is level and that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

When an object is in a circular motion, the magnitude of the net force can be found in terms of the mass and speed of that object, as well as the radius of the circular path:

`\begin{aligned} (\text{net force}) &= \frac{(\text{mass}) \, (\text{speed})^(2)}{(\text{radius})} \end{aligned}`.

Let
`m` denote the mass of this vehicle. Let
`v` denote the speed of this vehicle. Let
`r` denote the radius of this circular path. While in this circular motion, the magnitude of the net force will be on this vehicle will be:

`\begin{aligned} (\text{net force}) &= \frac{(\text{mass}) \, (\text{speed})^(2)}{(\text{radius})} = (m\, v^(2))/(r) \end{aligned}`.

Forces on this vehicle include:

• Gravitational attraction from the earth (weight) of magnitude
  `m\, g`, pointing downwards,
• Normal force from the road, pointing upwards, and
• Static friction from the road, pointing towards the center of the curve.

If this road is level, the normal force from the road will balance the weight of this vehicle. The magnitude of the normal force will be equal to that of the weight of this vehicle,
`m\, g`.

Let
`\mu_{\text{s}}` denote the static friction coefficient between the tire and the road. The static friction between the vehicle and the road cannot exceed
`(\text{static friction coefficient}) * (\text{normal force})`.

Since
`(\text{normal force}) = m\, g`, the maximum possible value of static friction will be:

`\begin{aligned} & (\text{static friction coefficient}) * (\text{normal force}) \\ =\; & (\mu_{\text{s}})\, (m\, g) \\ =& \mu_{\text{s}}\, m\, g\end{aligned}`.

Under the assumptions, the weight and normal force on this vehicle will be balanced. As a result, the net force on this vehicle will be equal to static friction and should also be no greater than
`(\text{static friction coefficient}) * (\text{normal force}) = \mu_{\text{s}}\, m\, g`.

In other words:

`\begin{aligned}(\text{net force}) &= (\text{static friction}) \\ &\le (\text{maximum static friction}) = \mu_{\text{s}}\, m\, g \end{aligned}`.

Additionally, from circular motion:

`\begin{aligned} (\text{net force}) &= (m\, v^(2))/(r) \end{aligned}`.

Therefore:

`\begin{aligned}(m\, v^(2))/(r) \le \mu_{\text{s}}\, m\, g \end{aligned}`.

Rearrange this inequality to separate the coefficient of static friction,
`\mu_{\text{s}}`:

`\begin{aligned} \mu_{\text{s}} \ge (v^(2))/(r\, g)\end{aligned}`.

(Note that
`m` and
`g` are both greater than
`0`.)

Substitute in
`v = 25.0\; {\rm m\cdot s^(-1)}`,
`r = 500\; {\rm m}`, and
`g \approx 9.81\; {\rm m\cdot s^(-2)}`:

`\begin{aligned} \mu_{\text{s}} &\ge (v^(2))/(r\, g) \\ & \approx \frac{(25.0\; {\rm m\cdot s^(-1)})^(2)}{(500\; {\rm m})\, (9.81\; {\rm m\cdot s^(-2)})} \\ &\approx (25.0^(2))/(500* 9.81) \\ &\approx 0.128\end{aligned}`.

(Rounded up.)