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.

Why the acceleration points inward

It can seem strange that an object moving at constant speed is accelerating, so it is worth being able to justify it. Velocity is a vector, and in circular motion its direction changes continuously even though its magnitude does not. Over a small time the velocity vector turns through a small angle, and the change in velocity points toward the centre of the circle. Acceleration is the rate of change of velocity, so it too points to the centre. A short geometric argument shows the magnitude of this change gives $a_c=v^2/r$. Because there is an inward acceleration, Newton's second law demands an inward net force, which is the centripetal force. This is the conceptual core examiners test with phrases like "explain why a body in uniform circular motion is accelerating".

Choosing the right form of the equation

The centripetal relations come in several equivalent forms, and picking the right one saves work. Use $a_c=v^2/r$ and $F_c=mv^2/r$ when you know the linear speed. Use $a_c=\omega^2 r$ when the angular speed is given or when the question is about rotation rate. If the period $T$ or frequency is given, substitute $v=\dfrac{2\pi r}{T}$ first, which turns the relation into $F_c=\dfrac{4\pi^2 m r}{T^2}$. For a satellite or a car the force itself is supplied by gravity or friction, so you then equate that real force to the chosen form and solve. Writing down which quantity you are given before choosing the form prevents the common waste of converting back and forth between $v$ and $\omega$.

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.