6.3 Centripetal force  (Page 3/15)

[link] shows a free-body diagram for a car on a frictionless banked curve. If the angle $\theta$ is ideal for the speed and radius, then the net external force equals the necessary centripetal force. The only two external forces acting on the car are its weight $\overrightarrow{w}$ and the normal force of the road $\overrightarrow{N}.$ (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude $m{v}^2\text{/}r.$ Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, so this must equal the centripetal force, that is,

Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From [link] , we see that the vertical component of the normal force is $N\text{cos}\theta ,$ and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,

Now we can combine these two equations to eliminate N and get an expression for $\theta$ , as desired. Solving the second equation for $N=mg\text{/}\left(cos\theta \right)$ and substituting this into the first yields

Taking the inverse tangent gives

This expression can be understood by considering how $\theta$ depends on v and r . A large $\theta$ is obtained for a large v and a small r. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve were frictionless. Note that $\theta$ does not depend on the mass of the vehicle.

Airplanes also make turns by banking. The lift force, due to the force of the air on the wing, acts at right angles to the wing. When the airplane banks, the pilot is obtaining greater lift than necessary for level flight. The vertical component of lift balances the airplane’s weight, and the horizontal component accelerates the plane. The banking angle shown in [link] is given by $\theta$ . We analyze the forces in the same way we treat the case of the car rounding a banked curve.