What is the maximum speed with which a 1140 kg car can round a turn of radius 65 m on a flat road if the coefficient of friction between tires and road is 0.80?
force friction= mu*mg
force centripetal= mv^2/r
set them equal, solve for v
To determine the maximum speed at which a car can round a turn, we can use the concept of centripetal force. The centripetal force required to keep the car moving in a circle is provided by the friction force between the tires and the road.
The formula for centripetal force is given by:
F = m * v^2 / r
Where:
F is the centripetal force (provided by the friction force),
m is the mass of the car (1140 kg),
v is the velocity of the car, and
r is the radius of the turn (65 m).
We can rearrange the formula to solve for v:
v^2 = F * r / m
v = √(F * r / m)
To find the maximum speed, we need to determine the maximum value for the friction force, which can be calculated using the coefficient of friction (μ) and the normal force (N) between the car and the road.
The formula for friction force is:
F_friction = μ * N
The normal force can be calculated as the product of the mass of the car and the acceleration due to gravity (g = 9.8 m/s^2), assuming the car is on a flat road:
N = m * g
Substituting the values into the equation, we get:
F_friction = μ * m * g
Now we can substitute this expression for friction force (F_friction) into the equation for velocity (v):
v = √(F * r / m)
v = √((μ * m * g) * r / m)
v = √(μ * g * r)
Let's plug in the given values:
μ = 0.80 (coefficient of friction)
m = 1140 kg (mass of the car)
r = 65 m (radius of the turn)
g = 9.8 m/s^2 (acceleration due to gravity)
v = √(0.80 * 9.8 * 65)
Calculating this expression, we find that the maximum speed at which a 1140 kg car can round a turn of radius 65 m on a flat road, with a coefficient of friction between tires and road of 0.80, is approximately 27.1 m/s.