A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
Practice Questions
1 question
Q1
A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
0.1
0.2
0.3
0.4
The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.
Questions & Step-by-step Solutions
1 item
Q
Q: A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
Solution: The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.
Steps: 10
Step 1: Understand that the car is moving in a circle, which means it needs a force to keep it moving in that circular path. This force is called centripetal force.
Step 2: The formula for centripetal force (F) is F = mv^2/r, where m is the mass of the car, v is the speed of the car, and r is the radius of the circular track.
Step 3: Identify the values given in the problem: the radius (r) is 100 m and the maximum speed (v) is 20 m/s.
Step 4: The frictional force that prevents the car from skidding is given by the formula F_friction = μmg, where μ is the coefficient of friction, m is the mass of the car, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 5: Set the centripetal force equal to the frictional force: mv^2/r = μmg.
Step 6: Notice that the mass (m) cancels out from both sides of the equation, simplifying it to v^2/r = μg.
Step 7: Rearrange the equation to solve for μ: μ = v^2/(rg).
Step 8: Substitute the known values into the equation: μ = (20^2)/(100 * 9.8).
