Circular Motion
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Direct answer
Uniform circular motion requires a net radial force of magnitude mv^2/r = m omega^2 r, directed toward the centre; the velocity stays tangential. The suppliers of this centripetal force are familiar forces in disguise: tension in a conical pendulum, the horizontal component of the normal reaction on a banked road where tan(theta) = v^2/(rg), friction on a level curve (mu > v^2/(rg) to avoid skidding), or gravity at the top of a vertical circle where the minimum speed on a string is sqrt(g r). Angular speed links them: omega = v/r = 2 pi n, with period T = 2 pi r/v. Centrifugal force is not a new force — it is the pseudo-force seen in the rotating frame.
What you must remember
- Kinematic links: v = omega r; a(c) = v^2/r = omega^2 r; T = 2 pi/omega; all three forms of centripetal acceleration appear in Main numericals.
- Level curve limit: on flat ground, friction alone turns the vehicle, so v(max) = sqrt(mu r g); doubling speed quadruples the friction demanded.
- Banking without friction: tan(theta) = v^2/(rg); at the optimum speed no friction is needed at all, and both tyres and road last longer — the engineering reason curves are banked.
- Banking with friction: maximum safe speed v(max) = sqrt(r g (tan(theta) + mu)/(1 − mu tan(theta))); the minimum speed formula swaps the signs.
- Conical pendulum: tan(theta) = v^2/(rg) for the string angle, and the period T = 2 pi sqrt(L cos(theta)/g) — smaller the angle, longer the period.
- Vertical circle, string or track: v(top) at least sqrt(gr); v(bottom) at least sqrt(5gr); T(bottom) − T(top) = 6 mg; at the top T + mg = mv^2/r, at the bottom T − mg = mv^2/r.
- Radius of curvature: for a projectile at the top of its flight, r = v(x)^2/g = u^2 cos^2(theta)/g — the direction where gravity acts radially.
- Pattern note: Main tests banked-road and vertical-circle numericals as single-correct; Advanced prefers the vertical circle on a rod, or motion leaving a track and switching to projectile motion.
One loop, three checkpoints
Take a ball sliding on the inside of a smooth vertical loop of radius r and ask what release height h on the adjoining track gets it around. Work in energy from release point to top: mg(h − 2r) = (1/2)mv(top)^2. Contact with the track demands v(top)^2 at least gr, so h at least 2.5r — the famous five-halves rule, the same arithmetic as the string pendulum's sqrt(5gr) at the bottom. If h lies between 2r and 2.5r, the ball reaches the upper region with too little speed, the normal reaction hits zero while the radius still points upward, and the ball departs the surface as a projectile — the follow-up question Advanced always appends, asking where it lands or with what angle it leaves.
The banked road deserves the same checkpoint habit. A curve of radius 100 m banked at 30 degrees has an optimum speed sqrt(rg tan(theta)) = sqrt(100 × 9.8 × 0.577) = 23.8 m/s — below it friction acts up the slope, above it down the slope, since friction responds to the deficiency or excess of speed.
Where students slip
Centrifugal "force" heads the list: it exists only in the rotating frame, and inserting mv^2/r as an outward real force in an inertial-frame equation produces equations that fail on contact with reality. The second error is applying sqrt(gr) at the wrong point of the vertical circle — it is the top speed for a string or outer track, not the bottom, and the bottom demands sqrt(5gr). The third is forgetting that on a rod (rigid constraint) no minimum top speed exists, since the rod can pull as well as push.
Frequently asked questions
Why is there no outward force on a body in uniform circular motion?
In an inertial frame the only real forces are toward the centre; the apparent outward pull is the centrifugal pseudo-force that appears only in the rotating frame of the body.
What speed allows a car to turn on a banked road without friction?
v = sqrt(r g tan(theta)), obtained by resolving the normal reaction so its horizontal component supplies the entire centripetal force.
What is the minimum speed at the top of a vertical circle on a string?
sqrt(gr), where gravity alone provides the centripetal acceleration and tension reduces to zero; any slower and the string slackens.
How do the tensions at the bottom and top of a vertical circle differ?
T(bottom) − mg = mv^2/r and T(top) + mg = mv^2/r; using energy between the two points, T(bottom) − T(top) = 6 mg.
What is the radius of curvature of a projectile's path at its highest point?
r = u^2 cos^2(theta)/g, because the horizontal speed u cos(theta) is the tangential speed there and g acts perpendicular to it.