A banked curve of radius Rin a new highway is designed so that a car traveling at speed vo can negotiate the turn safely on glare ice (zero friction). If a car travels too slowly.then it will slip toward the center of the circle. If it travels too fast, it will slip away from the center of the circle. If the coefficient of static friction increases, it becomes possible for a car to stay on the road while traveling at a speed within a range from Umin to Umax Part A Derive formula for v_(min)as a function of mu _(s),v_(0) and R Ekpress your answer in terms of the variables mu _(s) vo, R, and appropriate constants. sqrt [square ](square ) Asum d ? x^a x_(b) (a)/(b) sqrt (x) sqrt [n](x) x hat (x) vert xvert xcdot 10^n x v_(min)=sqrt ((v_(0)^2-mu _(s)gR))

To derive the formula for \( v_{min} \) as a function of \( \mu_s \), \( v_0 \), and \( R \), we need to consider the forces acting on the car as it negotiates the banked curve. The car is traveling at a speed \( v \) and the radius of the curve is \( R \). The centripetal force required to keep the car moving in a circular path is provided by the horizontal component of the normal force and the frictional force.<br /><br />The forces acting on the car are:<br />1. The gravitational force \( mg \) acting vertically downward.<br />2. The normal force \( N \) acting perpendicular to the surface of the banked curve.<br />3. The frictional force \( f \) acting parallel to the surface of the banked curve.<br /><br />The centripetal force \( F_c \) required for circular motion is given by:<br />\[ F_c = \frac{mv^2}{R} \]<br /><br />On a banked curve, the horizontal component of the normal force provides part of this centripetal force. If there is no friction (\( \mu_s = 0 \)), the entire centripetal force must be provided by the horizontal component of the normal force. Therefore, we have:<br />\[ N \sin(\theta) = \frac{mv^2}{R} \]<br />where \( \theta \) is the angle of the banked curve.<br /><br />The normal force \( N \) is also given by:<br />\[ N = mg \cos(\theta) \]<br /><br />Combining these two equations, we get:<br />\[ mg \cos(\theta) \sin(\theta) = \frac{mv^2}{R} \]<br />\[ g \sin(\theta) \cos(\theta) = \frac{v^2}{R} \]<br />\[ g \sin(\theta) = \frac{v^2}{R} \]<br />\[ v^2 = gR \tan(\theta) \]<br /><br />Now, let's consider the case where there is friction. The frictional force \( f \) acts to oppose the motion of the car and is given by:<br />\[ f = \mu_s N \]<br /><br />The total centripetal force is now provided by the horizontal component of the normal force and the frictional force:<br />\[ N \sin(\theta) + \mu_s N \cos(\theta) = \frac{mv^2}{R} \]<br /><br />Substituting \( N = mg \cos(\theta) \) into the equation, we get:<br />\[ mg \cos(\theta) \sin(\theta) + \mu_s mg \cos(\theta) \cos(\theta) = \frac{mv^2}{R} \]<br />\[ g \sin(\theta) \cos(\theta) + \mu_s g \cos(\theta) \cos(\theta) = \frac{v^2}{R} \]<br />\[ g \sin(\theta) + \mu_s g \cos(\theta) = \frac{v^2}{R} \]<br />\[ v^2 = gR \left( \sin(\theta) + \mu_s \cos(\theta) \right) \]<br /><br />To find the minimum speed \( v_{min} \) at which the car can stay on the road, we need to ensure that the car does not slip. This happens when the frictional force is zero, i.e., \( \mu_s = 0 \). Therefore, the minimum speed \( v_{min} \) is given by:<br />\[ v_{min} = \sqrt{gR \sin(\theta)} \]<br /><br />However, since \( \theta \) is the angle of the banked curve, we can use the relationship \( \sin(\theta) = \frac{v_0^2}{gR} \) to express \( v_{min} \) in terms of \( v_0 \), \( \mu_s \), and \( R \):<br />\[ v_{min} = \sqrt{gR \left( \frac{v_0^2}{gR} + \mu_s \right)} \]<br />\[ v_{min} = \sqrt{v_0^2 + \mu_s gR} \]<br /><br />Therefore, the formula for \( v_{min} \) is:<br />\[ v_{min} = \sqrt{v_0^2 + \mu_s gR} \]