Centrifugal Force Calculator

What the centrifugal force is

A rock on a string moving in a circle at a steady speed is still accelerating, because its direction keeps changing. Watch it from the ground and one force explains the motion: the string pulls the rock toward your hand with a force of m v squared / r. That inward pull is the centripetal force, and it is the only real force acting on the rock.

Now sit on the rock. In that frame the rock is not moving at all, yet the string still pulls inward. To keep the equation force = mass x acceleration true when the acceleration reads zero, you add an outward force of the same size. That term is the centrifugal force. Nothing is pushing outward. The term exists to pay for using a frame that is turning.

NASA's Goddard Space Flight Center teaching page on frames of reference, written by David P. Stern and last updated in 2004, states the rule in the same terms: equilibrium exists in the rotating frame only if all forces balance, including an inertial centrifugal force of m v squared / r directed outward along the radius.

Both common sentences are therefore correct. The outward pull is what your body feels, and no object is applying it. The only interaction is the inward one.

The formula in two forms

Two versions of the same expression cover almost every case:

F = m v squared / r, where v is the speed along the circle, and F = m w squared r, where w is the angular speed in radians per second. The second form follows from v = w r.

Units decide whether the answer is right. Mass in kilograms, speed in meters per second and radius in meters give newtons, because the newton is defined as a kilogram meter per second squared (NIST Special Publication 811, 2008, Table 3). Feed in grams, centimeters or kilometers per hour and the number that comes out is not a force.

Angular speed needs one conversion. A full turn is 2 pi radians, so one turn per second is 6.28319 radians per second, and a motor rated in revolutions per minute converts as w = rpm x 2 pi / 60 = rpm x 0.10472.

The acceleration is a = v squared / r = w squared r, and the g multiple is that value divided by 9.80665 meters per second squared, the standard acceleration of free fall. The direction of the acceleration is toward the axis, which is the part people flip by accident.

Example: 0.2 kg on a 0.5 m string

A 0.2 kg mass on a 0.5 m string turns 3 times per second, which is 180 rpm. Work it in the order the units arrive.

Both forms land on 35.5306 newtons, which is 7.99 pounds-force. The acceleration is 88.82644 / 0.5 = 177.65 meters per second squared, or 18.12 g. For a sense of scale, 35.53 newtons is about the weight of an 8 pound bag sitting still, so a light mass on a short string at three turns a second already pulls like a bag of rice.

The check line is worth keeping. If the two forms disagree, the radius or the speed was converted twice.

Example: a centrifuge drum at 1,200 rpm

A rotor holds 1.5 kg of sample at a radius of 0.25 m and spins at 1,200 rpm. Convert the spin first: 1,200 x 2 pi / 60 = 125.6637 radians per second. Squared, that is 15,791.37. Multiply by the radius: 15,791.37 x 0.25 = 3,947.84 meters per second squared. Divide by 9.80665 and the load sits at 402.57 g. Multiply by the mass and the force is 5,921.76 N, or 1,331.27 pounds-force.

That last number explains the hardware. A 1.5 kg load pulling with 1,331 pounds of force is why a benchtop centrifuge has a lid interlock and why an unbalanced tube walks the machine across the bench.

Read the table by column, not by row. Going from 1,000 to 2,000 rpm quadruples the g, because the speed is squared. Doubling the radius at a fixed spin only doubles it.

Unit traps that change the answer

• rpm against radians per second. At a 0.25 m radius, 1,200 rpm is 402.57 g. Feed 1,200 in as radians per second instead and you get 36,710 g, 91 times too high. The two units differ by a factor of 9.55, and the force scales with its square.
• Speed. 100 km/h is 27.78 m/s and 60 mph is 26.82 m/s. A speed left in km/h makes the force about 13 times too large.
• Radius. 25 cm is 0.25 m, 10 inches is 0.254 m and 1 foot is 0.3048 m. Radius is measured to the axis of rotation, never to the outside of the drum.
• Pounds. The pound is a mass (0.45359237 kg) and the pound-force is a force (4.4482216152605 N). A 5 lb mass is 2.268 kg, and at 400 g it pulls with 8,896.6 N, near 2,000 lbf.

The last row is the one to remember when a spec sheet talks in pounds. Divide newtons by 4.4482216152605, or multiply by 0.2248089, and the answer is a force your reader can picture.

Where the number shows up

Laboratory centrifuges are rated in g rather than rpm for a reason: the same spin gives different g on different rotors. At 1,200 rpm a 0.25 m rotor produces 402.57 g, while a 0.5 m rotor produces 805.14 g at the same speed. A protocol that says 400 g is portable between machines; one that says 1,200 rpm is not.

A rotating space station uses the same arithmetic in reverse. To make 1 g at a 100 m radius, w = the square root of (9.80665 / 100) = 0.31315 radians per second, which is 2.99 turns per minute. The 2007 NASA chapter on the physics of artificial gravity notes that the force grows with distance from the axis, so a crew member's feet feel more than their head in a short-radius design.

Rides, rotating machinery and flywheels are sized the same way. A car turning a corner is usually analyzed in the frame of the road, where friction is the inward force and no outward force appears at all. That case belongs to the centripetal calculator, and the numbers match this one whenever the motion is steady.

Where the formula stops

The formula treats the object as a point at one radius. A long object spans a range of radii, so each end feels a different force, and the single number is only an average. NASA's artificial gravity chapter calls this the gravity gradient, and it is why short-radius centrifuges are uncomfortable.

It also assumes a steady spin. Change the speed and an extra term appears along the direction of motion, and move something inside the rotating frame and a Coriolis term appears at right angles to the motion. Neither changes the size of the centrifugal term, but both change what a person inside the frame does feel.

Last, the force is frame-dependent. In the ground frame it is exactly zero, and the inward force does all the work. Switch frames and the bookkeeping changes while the physical situation does not. A calculation that mixes the two, adding a centrifugal force to a ground-frame free body diagram, will double count and come out wrong.