Question

27. A car travels along a horizontal road which is an
arc of a circle of radius 125m. the greatest speed.
at which the car can travel without slipping is
42km/hr. Find the coefficient of friction between
the tyres of the car and the road surface.

Answer 1

To start solving this problem, we need to use the formula for centripetal force:

F = m * (v^2 / r)

Where:
F is the centripetal force
m is the mass of the car
v is the velocity of the car
r is the radius of the circle

Since the car is not slipping, the frictional force provides the necessary centripetal force. Therefore, we can write:

F_friction = m * (v^2 / r)

The frictional force can also be written as:

F_friction = μ * N

Where:
μ is the coefficient of friction
N is the normal force

The normal force, N, is equal to the weight of the car, so we can write:

N = m * g

Where:
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Substituting these equations into our original equation, we get:

μ * m * g = m * (v^2 / r)

Simplifying further:

μ * g = v^2 / r

Now we can plug in the given values:

g ≈ 9.8 m/s^2
v = 42 km/hr = (42 * 1000) / 3600 m/s (converting km/hr to m/s)
r = 125 m

Calculating further:

μ * 9.8 = ((42 * 1000) / 3600)^2 / 125

μ * 9.8 = 441 / 125

μ = (441 / 125) * (1 / 9.8)

μ ≈ 0.36

Therefore, the coefficient of friction between the tires of the car and the road surface is approximately 0.36.