Apply centripetal force and acceleration to horizontal, banked and vertical circular motion

What this dot point is asking

WACE wants you to see that "centripetal force" is not a new force but a role: whatever real force points toward the centre provides the net inward force needed to curve the path. Constant speed does not mean constant velocity, because velocity is a vector and its direction keeps changing.

Centripetal acceleration and force

For speed $v$ around a circle of radius $r$,

where the angular speed $\omega=2\pi/T$ and $T$ is the period. The acceleration points to the centre, so the net force must too. There is no outward "centrifugal force" in this frame; the outward feeling is inertia resisting the inward pull.

Horizontal circles

For a ball on a string swung in a horizontal circle, the horizontal component of the string tension supplies $F_c$ while the vertical component balances weight. For a car on a flat road, static friction between the tyres and road provides the entire centripetal force, so the maximum cornering speed is set by $\mu mg=mv^2/r$.

Banked tracks

On a frictionless banked track of angle $\theta$, the horizontal component of the normal force supplies $F_c$ while the vertical component balances weight:

Dividing gives the design speed for the bank,

At this speed no friction is needed at all.

Vertical circles

In a vertical circle the speed varies, so it is not strictly uniform, but the centripetal relationship still holds instant by instant. At the top, gravity and the normal force (or tension) both point inward:

The minimum speed to keep contact at the top is found by setting $N=0$, giving $v_{\min}=\sqrt{gr}$. At the bottom the support force must exceed weight to bend the path upward.

Naming the real force

In any answer, first state which physical force points toward the centre, then equate it to $mv^2/r$. Examiners reward this identification heavily, because it shows you understand that the centripetal force is supplied, not added.