A car travels around a circular turn of radius 50 meters while maintaining a constant speed of 15m/s
find the minimum value for the coefficient of friction necessary to keep the car on the road
in icy conditions, the coefficient of friction drops to 0.15, Now find the maximum safe speed for taking the turn.
mv²/R = μmg
(a) μ=v²/Rg = …
(b) v₁=sqrt(μ₁Rg)
To find the minimum value for the coefficient of friction necessary to keep the car on the road, we can start by considering the forces acting on the car as it takes the turn.
The car experiences two forces: the centripetal force, which keeps the car moving in a circle, and the frictional force, which prevents the car from slipping off the road.
The centripetal force is given by the equation: Fc = (mv^2)/r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the turn.
The frictional force can be calculated using the equation: Ff = μN, where μ is the coefficient of friction and N is the normal force acting on the car. In this case, the normal force is equal to the car's weight, which is given by the equation: N = mg, where g is the acceleration due to gravity.
For the car to stay on the road, the frictional force must be equal to or greater than the centripetal force. Therefore, we can set up the following inequality:
μN >= (mv^2)/r
Substituting the expressions for N and mv^2 from above, we get:
μmg >= (mv^2)/r
The mass cancels out on both sides, and we are left with:
μg >= v^2/r
Now, plugging in the given values, we have:
μ * 9.8 >= (15^2) / 50
μ * 9.8 >= 225 / 50
μ * 9.8 >= 4.5
Dividing both sides by 9.8, we find:
μ >= 4.5 / 9.8
μ >= 0.46
Therefore, the minimum value for the coefficient of friction necessary to keep the car on the road is approximately 0.46.
Now, to find the maximum safe speed for taking the turn in icy conditions, given that the coefficient of friction is 0.15, we can rearrange the inequality we derived earlier:
μg >= v^2/r
0.15 * 9.8 >= v^2 / 50
1.47 >= v^2 / 50
Multiplying both sides by 50, we get:
73.5 >= v^2
Taking the square root of both sides, we find:
v <= √73.5
v <= 8.58 m/s
Therefore, the maximum safe speed for taking the turn in icy conditions is approximately 8.58 m/s.