Question

A turn of radius 20 m is banked for the vehicles going
at a speed of 36 km/h. If the coefficient of static friction
between the road and the tyre is 0.4, what are the
possible speeds of a vehicle so that it neither slips down
nor skids up ?

Answer 1

Answer on Question #58067, Physics / Mechanics | Relativity

A turn of radius $20\mathrm{m}$ is banked for the vehicles going at a speed of $36~\mathrm{km / h}$ . If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Solution:

The net force must be horizontal--pointing toward the center of the circle--and only the friction force is available to provide it. The normal force and the weight simply cancel each other.

The centripetal force,

$$F _ {c} = \frac {M v ^ {2}}{r}$$

For this flat curve, the centripetal force is supplied by the friction force, $F_{f}$ ,

$$F _ {f} = \mu N = \mu M g$$ $$F _ {f} = F _ {c}$$

Thus,

$$\frac {M v ^ {2}}{r} = \mu M g$$ $$v = \sqrt {\mu g r} = \sqrt {0 . 4 \cdot 9 . 8 \cdot 2 0} = 8. 8 5 \frac {\mathrm {m}}{\mathrm {s}} = 8. 8 5 \cdot 3. 6 \frac {\mathrm {k m}}{\mathrm {h}} \approx 3 1. 9 \frac {\mathrm {k m}}{\mathrm {h}}$$

Answer: $31.9 \frac{\mathrm{km}}{\mathrm{h}}$

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