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.
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.
1 answer
1 answer