Circular motion

What the exam asks

Expect a car on a bend, a mass on a string, a conical pendulum, or a body at the top of a loop, and a question asking which force provides the centripetal force or how a change in speed or radius alters it. The commonest items are: pick the correct free body diagram, decide whether mass affects the answer, and reason about what happens at the limit when friction or tension runs out. The trap that catches most candidates is drawing centripetal force as its own arrow alongside the friction or tension that is supplying it, which double counts one force and gives an answer twice too large. Ask what is touching the body, resolve those forces towards the centre, and set the sum equal to mv squared over r. The second trap is centrifugal force, which appears in a wrong option in almost every deck of this topic and is never the answer to what is acting on the body.

A body moving in a circle at steady speed is accelerating, because its direction is changing, and the acceleration points at the centre. That acceleration is v squared over r, and by F = ma the force needed to sustain it is mv squared over r. Both are supplied in the stem when a question needs them.

The idea the exam is actually testing is different, and it is the one thing worth carrying in: centripetal force is a role, not a new force. Nothing in nature is a centripetal force. It is a job that some force already in the free body diagram has to do, and the whole question is usually naming which one. For a car on a flat bend it is friction. On a frictionless banked track it is the horizontal component of the normal force. For a conical pendulum it is the horizontal component of the tension. For a planet it is gravity.

So the method is: draw the free body as you would for any statics problem, resolve towards the centre, and set that sum equal to mv squared over r. Do not add an arrow labelled centripetal, because you would be counting a force twice, and never add one labelled centrifugal, because in the ground frame there is nothing touching the body that pushes it outwards. What feels like an outward push in a turning car is the body's inertia continuing straight while the car turns underneath it.

What to hold

• Circular motion at constant speed is accelerated motion, because velocity is a vector and its direction is changing continuously.
• The centripetal acceleration is v squared over r and points at the centre of the circle, so the net force must point there too.
• Centripetal force is a role played by a force that is already acting, not an additional force. Identifying which force plays it is usually the whole question.
• On a flat curve the centripetal force is friction between the tyres and the road, and there is nothing else available to supply it.
• On a frictionless banked curve it is the horizontal component of the normal force, which is why banking lets a car turn on ice in principle.
• For a conical pendulum it is the horizontal component of the string tension, while the vertical component supports the weight.
• Centrifugal force does not belong on a free body diagram drawn in the ground frame, because no object is touching the body and pushing it outwards.
• The outward feeling in a turning car is inertia: the passenger continues in a straight line while the car curves, so the door comes to meet them.
• The maximum speed on a flat curve is the square root of μrg, which contains no mass, so a loaded truck and an empty one slide at the same speed.
• The banking angle satisfies tan θ = v squared over rg, which also has no mass in it, so one angle serves every vehicle at that speed.
• Centripetal force does no work, because it is always perpendicular to the velocity, which is why speed is constant even though the body is accelerating.
• Force scales with the square of speed and inversely with radius, so doubling the speed quadruples the force needed and halving the radius doubles it.
• Cut the string and the body flies off along the tangent, not radially outward, because with no force acting it continues in the straight line it was already travelling.