1. System Analysis and Diagram

Consider the car moving in a circular path of radius $R$ on a plane inclined at an angle $\theta$. Unlike a banked turn where the road is banked for the turn, here the entire parking lot is a slope, and the car drives in a circle on this slope.

We define the forces acting on the car. The critical point for slipping occurs at the lowest point of the circular path. At this point, the center of the circle is directly “up” the slope.

2. Force Equations

The car performs circular motion on the plane. The centripetal force $\vec{F}_{cp}$ must be directed towards the center of the circle. At the lowest point of the circle (closest to the bottom of the slope), the center $C$ is located directly up the slope.

The forces acting on the car along the line of the slope are:

• Component of Gravity: $mg \sin \theta$ acting down the slope (away from the center).
• Static Friction ($f$): To prevent slipping and provide the necessary centripetal acceleration, friction must act up the slope (towards the center).

The equation for the net radial force providing the centripetal acceleration is: $$ f – mg \sin \theta = \frac{mv^2}{R} $$

Solving for the friction required: $$ f = \frac{mv^2}{R} + mg \sin \theta $$

3. Condition for Maximum Speed

The car slips when the required friction exceeds the maximum static friction available. The maximum static friction is given by: $$ f_{\text{max}} = \mu N $$ Since the plane is inclined at $\theta$, the normal reaction is $N = mg \cos \theta$. Thus: $$ f_{\text{max}} = \mu mg \cos \theta $$

To find the greatest speed $v_{\text{max}}$ without slipping, we set the required friction equal to the limiting friction: $$ \frac{mv_{\text{max}}^2}{R} + mg \sin \theta = \mu mg \cos \theta $$

4. Solving for Velocity

Rearranging the equation to solve for $v_{\text{max}}$:

$$ \frac{mv_{\text{max}}^2}{R} = \mu mg \cos \theta – mg \sin \theta $$

Cancel the mass $m$ from both sides: $$ \frac{v_{\text{max}}^2}{R} = g(\mu \cos \theta – \sin \theta) $$

Multiply by $R$ and take the square root: $$ v_{\text{max}}^2 = gR(\mu \cos \theta – \sin \theta) $$ $$ v_{\text{max}} = \sqrt{gR(\mu \cos \theta – \sin \theta)} $$

Note: This condition is derived at the bottom of the loop where the demand on friction is highest (opposing gravity). If the car does not slip here, it will not slip at other points where gravity assists the centripetal force or acts perpendicular to it.