A 1220 kg car rounds a circular turn of radius 19.2 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.66, how fast can the car go without skidding?
Let mu_s be the static friction coefficient.
If the road is not banked and the car is on the verge of slipping, the maximum static friction force equals the centripetal force.
M*g*mu_s = M*V^2/R
Cancel the M's.
V^2 = R*g*0.66
Solve for V
5.6
isn't it 11.14? How the heck did you end up with 5.6?
V = sqrt(19.2 * 9.8 * 0.66)
V = 11.14
11.14, that is
To determine how fast the car can go without skidding, we need to find the maximum speed at which the inward frictional force can provide the necessary centripetal force for the car to stay in circular motion.
The centripetal force is given by the equation:
F_c = m * a_c,
where F_c is the centripetal force, m is the mass of the car, and a_c is the centripetal acceleration.
The centripetal acceleration can be calculated using the equation:
a_c = v^2 / r,
where v is the velocity of the car and r is the radius of the circular turn.
Since the centripetal force is provided by the static friction between the tires and the road, we can write:
F_c = f_s,
where f_s is the static frictional force.
From the equation for static friction:
f_s = μ_s * N,
where μ_s is the coefficient of static friction and N is the normal force.
For a car on a flat road, the normal force is equal to the weight of the car, which can be calculated as:
N = m * g,
where g is the acceleration due to gravity.
Combining the equations:
m * a_c = μ_s * m * g.
Canceling out the mass:
a_c = μ_s * g.
Substituting for a_c, we have:
v^2 / r = μ_s * g.
Solving for v, we get:
v = sqrt(μ_s * g * r).
Now we can plug in the given values:
μ_s = 0.66 (coefficient of static friction),
g = 9.8 m/s^2 (acceleration due to gravity),
r = 19.2 m (radius of the circular turn).
Substituting these values into the equation, we get:
v = sqrt(0.66 * 9.8 * 19.2).
Calculating this, we find:
v ≈ 17.8 m/s.
Therefore, the car can go at a maximum speed of approximately 17.8 m/s without skidding.