A banked circular highway curve is designed for traffic moving at \(95 \mathrm{~km} / \mathrm{h}\). The radius of the curve is \(210 \mathrm{~m}\). Traffic is moving along the highway at \(52 \mathrm{~km} / \mathrm{h}\) on a stormy day. ( \(a\) ) What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding? \((b)\) With this value of the coefficient of friction, what is the greatest speed at which the curve can be negotiated without sliding?

Imagine you're trying to push a heavy box across the floor. The force you have to overcome before the box starts sliding is a direct result of the coefficient of friction between the box and the floor. In physics, this coefficient is a dimensionless number that represents the ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing those two surfaces together.

When it comes to banked curve problems in physics, the coefficient of friction between the car's tires and the road determines how well the car can maneuver the curve without sliding off. During a stormy day, as described in our exercise, the road becomes slippery, lowering this valuable coefficient. To ensure safety, civil engineers must calculate the minimum coefficient of friction that will allow for safe travel even under adverse weather conditions. The formula used in such a scenario is \[\mu = \frac{v^2}{rg} - \tan\theta\]where \(v\) is the velocity of the car, \(r\) is the radius of the curve, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of banking. If \(\theta\) is zero—representing a flat road—the formula simplifies even further.

Have you ever tied a ball to a string and spun it around in a circle? The force that keeps the ball moving in a circular path and not flying off in a straight line is called the centripetal force. In a banked curve scenario on a highway, this force is necessary to keep the vehicle on a curved path rather than continuing straight due to inertia. For circular motion to occur, there must always be a centripetal force directed towards the center of the circle.

The centripetal force formula in our case is \[F_c = mv^2/r\]where \(m\) is the mass of the vehicle, \(v\) is the velocity, and \(r\) is the radius of the highway curve. As identified in step one of the solution, without sufficient friction—or in the absence of it—the only source of centripetal force is the horizontal component of the reaction from the banked road. On a flat road, if the velocity of the vehicle is too high, the car may need the additional frictional force to provide the extra centripetal force necessary to stay on the curve.

Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform, with a changing rate of rotation.

In the context of banked curves, like the one in our textbook exercise, we deal with uniform circular motion. This is where the speed of the vehicle—and consequently its centripetal acceleration—remains constant. When a car travels around a banked curve, the forces involved must be balanced in order for the car not to slip or skid. This means that the centripetal force required for circular motion must equal the force provided by the static friction and any component of gravitational force acting in the horizontal direction if the road is banked.

For vehicles on a banked curve during a storm, it is critical that the available frictional force, combined with the natural centripetal force due to the road's banking, is strong enough to maintain circular motion. Otherwise, the vehicle risks sliding outwards off the curve due to insufficient lateral force to keep it on the intended circular path.