the cofficient of friction b/n the tyre of the car rotating around horizontal circuar road way is 0.5; what would the minimum radius at which acar turn ahorizontal road when its speed is 15m/s?g=10m/s
To determine the minimum radius at which a car can turn on a horizontal roadway, we first need to consider the centripetal force acting on the car as it rotates.
The centripetal force can be calculated using the formula:
Fc = mv^2 / r
Where:
Fc is the centripetal force
m is the mass of the car
v is the velocity of the car
r is the radius of the circular path
In this case, we are given that the coefficient of friction (μ) between the car's tires and the road is 0.5, and the car's speed (v) is 15 m/s, and the acceleration due to gravity (g) is 10 m/s^2.
The maximum frictional force (f) that the tires can provide can be calculated by:
f = μ * (mass of the car) * g
Given that g = 10 m/s^2 and μ = 0.5, we can calculate f:
f = 0.5 * (mass of the car) * 10
Now, the maximum frictional force (f) must provide the centripetal force (Fc) for the car to make the turn without slipping. Therefore, we can equate these two forces:
f = Fc
0.5 * (mass of the car) * 10 = (mass of the car) * v^2 / r
We can cancel out the mass of the car on both sides of the equation:
0.5 * 10 = v^2 / r
5 = v^2 / r
r = v^2 / 5
Substituting the given value of v = 15 m/s into the equation, we can calculate the minimum radius (r):
r = (15)^2 / 5
r = 225 / 5
r = 45 meters
Therefore, the minimum radius at which the car can turn on a horizontal roadway with a speed of 15 m/s and a coefficient of friction of 0.5 is 45 meters.