Question

A motorcyclist goes around an un-banked (i.e., flat) circular turn of radius 31m, at a constant speed of 110km/hr (convert this to m/s). What is the minimum coefficient of static friction needed to keep the tires from slipping? Explain why the answer is (or is not) plausible.

Answer 1

Answer:

$\mu_{s} = 3.071$

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Explanation:

From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:

$\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$N$ - Normal force, measured in newtons.

$m$ - Mass of the motocyclist, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Speed of the motorcyclist, measured in meters per second.

$R$ - Radius of the circular turn, measured in meters.

The static coefficient of friction is cleared in (Eq. 1):

$\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}$

From (Eq. 2) we get that normai force is:

$N = m\cdot g$

And we expand the resulting expression in (Eq. 1):

$\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$ (Eq. 3)

If we know that $v = 30.556\,\frac{m}{s}$, $g = 9.807\,\frac{m}{s^{2}}$ and $R = 31\,m$, the expected static coefficient of friction is:

$\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}$

$\mu_{s} = 3.071$

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.