Questions: What is the maximum speed (in m / s ) at which a car can round a corner with radius 25.0 m, if the coefficient of static friction between its tires and the road is 0.40?
Transcript text: Question 1. What is the maximum speed (in $\mathrm{m} / \mathrm{s}$ ) at which a car can round a corner with radius 25.0 m , if the coefficient of static friction between its tires and the road is 0.40 ?
Solution
Solution Steps
Step 1: Identify the forces involved
The maximum speed at which a car can round a corner is determined by the centripetal force, which is provided by the frictional force between the tires and the road. The frictional force can be calculated using the coefficient of static friction and the normal force.
Step 2: Write the equation for centripetal force
The centripetal force Fc required to keep the car moving in a circle of radius r at speed v is given by:
Fc=rmv2
where m is the mass of the car.
Step 3: Write the equation for frictional force
The maximum frictional force Ff that can act on the car is given by:
Ff=μsFn
where μs is the coefficient of static friction and Fn is the normal force. For a car on a flat road, the normal force Fn is equal to the weight of the car mg.
Step 4: Set the frictional force equal to the centripetal force
For the car to round the corner without slipping, the frictional force must be equal to or greater than the centripetal force:
μsmg=rmv2
Step 5: Solve for the maximum speed v
Cancel the mass m from both sides of the equation and solve for v:
μsg=rv2v2=μsgrv=μsgr
Step 6: Substitute the given values
Given:
μs=0.40
g=9.81m/s2
r=25.0m
Substitute these values into the equation:
v=0.40×9.81m/s2×25.0mv=98.1v≈9.905m/s