# Centripetal force, moment of inertia and torque calculator

> Centripetal force and acceleration from mass, speed and radius; rpm to rad/s and period; moment of inertia of nine shapes; torque from force and lever arm.

Interactive version: https://www.calcopenly.com/science/circular-motion-rotation-calculator
Subject: Science calculators

Centripetal force is the inward pull that keeps a mass on a circular path: F = mv²/r, with the centripetal acceleration a = v²/r = ω²r pointing at the centre. The other modes convert a rotation rate between rpm, rad/s, hertz and period (ω = 2πf), give the moment of inertia I of nine standard shapes, and work out torque τ = rF sin θ and angular acceleration α = τ/I.

The default is a 1,000 kg car taking a 50 m radius curve at 20 m/s (72 km/h). The tyres must supply 8,000 N of sideways grip, an acceleration of 8 m/s² or 0.816 g. In torque mode, a 50 N push at 90° on a 0.3 m wrench gives 15 N·m.

Results assume uniform circular motion and rigid bodies of uniform density. Mass farther from the axis raises I, so a thin hoop has twice the moment of inertia of a solid disc with the same mass and radius.

## Inputs

- **Calculate** (options: Centripetal force, Angular velocity, Moment of inertia, Torque)
- **Solve for** (options: Force, Speed, Radius)
- **Mass**
- **Speed**
- **Radius of the circle**
- **Centripetal force**
- **Rotation rate**
- **Mass of the body**
- **Shape and axis** (options: Point mass at radius R, Thin hoop or thin-walled cylinder (central axis), Solid cylinder or disc (central axis), Thick-walled cylinder (central axis), Solid sphere (through centre), Thin spherical shell (through centre), Thin rod about its centre, Thin rod about one end, Rectangular plate (axis through centre, perpendicular))
- **Radius R (outer R₁ for thick cylinder)**
- **Inner radius R₂**
- **Length L (side a for plate)**
- **Side b**
- **Spin rate (for rotational energy)**
- **Lever arm r (axis to point of force)**
- **Applied force**
- **Angle between arm and force**
- **Moment of inertia (for angular acceleration)**

## Results

- Centripetal force (N) — main result
- Centripetal acceleration (m/s²)
- Speed (m/s)
- Speed (km/h)
- Radius (m)
- Angular velocity (rad/s)
- Revolutions per minute (rpm)
- Frequency (Hz)
- Period (one revolution) (s)
- Moment of inertia (kg·m²)
- Rotational kinetic energy (J)
- Angular momentum (kg·m²/s)
- Torque (N·m)
- Angular acceleration (rad/s²)
- Centripetal acceleration (g)

## Formula

$$
a_c = \frac{v^2}{r} = \omega^2 r,\quad F_c = \frac{mv^2}{r},\quad \omega = 2\pi f = \frac{2\pi\,\text{rpm}}{60},\quad I = \sum m r^2,\quad \tau = rF\sin\theta = I\alpha
$$

## Worked examples

### 1000 kg car at 20 m/s on a 50 m curve

- Calculate: Centripetal force
- Solve for: Force
- Mass: 1000 kg
- Speed: 20 m/s
- Radius of the circle: 50 m
- **Centripetal force: 8,000 N**
- **Centripetal acceleration: 8 m/s²**
- **Angular velocity: 0.4 rad/s**
- Checked against: Python 3.8 decimal: a = 400/50 = 8, F = 8000, ω = v/r = 0.4

### Speed that needs 500 N on 10 kg at 2 m

- Calculate: Centripetal force
- Solve for: Speed
- Mass: 10 kg
- Radius of the circle: 2 m
- Centripetal force: 500 N
- **Speed: 10 m/s**
- Checked against: Python 3.8 decimal: v = √(Fr/m) = √100

### 3000 rpm at 10 cm

- Calculate: Angular velocity
- Radius of the circle: 0.1 m
- Rotation rate: 3000 rpm
- **Angular velocity: 314.159 rad/s**
- **Frequency: 50 Hz**
- **Period (one revolution): 0.02 s**
- **Speed: 31.4159 m/s**
- Checked against: Python 3.8 math: ω = 2π·3000/60, v = ωr

### Solid sphere 5 kg, R = 0.2 m

- Calculate: Moment of inertia
- Mass of the body: 5 kg
- Shape and axis: Solid sphere (through centre)
- Radius R (outer R₁ for thick cylinder): 0.2 m
- **Moment of inertia: 0.08 kg·m²**
- Checked against: Serway Table 10.2: I = 2/5 MR² = 0.4 × 5 × 0.04

### Rod about its end, 2 kg × 3 m, spinning at 60 rpm

- Calculate: Moment of inertia
- Mass of the body: 2 kg
- Shape and axis: Thin rod about one end
- Length L (side a for plate): 3 m
- Spin rate (for rotational energy): 60 rpm
- **Moment of inertia: 6 kg·m²**
- **Rotational kinetic energy: 118.435 J**
- **Angular momentum: 37.6991 kg·m²/s**
- Checked against: Python 3.8 math: I = ML²/3 = 6; ω = 2π; KE = ½Iω² = 12π² = 118.4353; L = 12π

### Wrench: 50 N at 90° on a 0.3 m arm

- Calculate: Torque
- Lever arm r (axis to point of force): 0.3 m
- Applied force: 50 N
- Angle between arm and force: 90 °
- Moment of inertia (for angular acceleration): 0.5 kg·m²
- **Torque: 15 N·m**
- **Angular acceleration: 30 rad/s²**
- Checked against: Python 3.8: τ = 0.3 × 50 × sin90° = 15; α = τ/I = 30

## Questions

### What is the formula for centripetal force?

F = mv²/r, with mass m in kilograms, speed v in metres per second and radius r in metres, giving newtons. Equivalent forms are F = mω²r and F = 4π²mr/T². Doubling the speed quadruples the force and doubling the radius halves it: a 1,000 kg car at 20 m/s on a 50 m curve needs 8,000 N, and at 40 m/s it would need 32,000 N.

### How do you convert rpm to rad/s?

Multiply rpm by 2π/60, about 0.10472. 3,000 rpm is 314.159 rad/s, or 50 revolutions per second (50 Hz), so one revolution takes 0.02 s. To go back, multiply rad/s by 60/(2π), about 9.5493. The radian is the coherent SI unit of plane angle (BIPM SI Brochure), so rad/s is the SI unit of angular velocity.

### Is centrifugal force real?

Not in an inertial, non-rotating frame of reference: there, the only horizontal force on a cornering car is the inward centripetal force from the tyres. Centrifugal force appears only when motion is described from inside the rotating frame, where it has the same size, mv²/r, pointing outward. The outward push a passenger feels is their body's inertia carrying it in a straight line while the car turns.

### What is the moment of inertia of a solid disc?

I = ½MR² about its central axis, so a 5 kg disc of radius 0.2 m has I = 0.5 × 5 × 0.2² = 0.1 kg·m². Other standard results are MR² for a thin hoop, ⅖MR² for a solid sphere, ⅔MR² for a thin spherical shell, and ML²/12 or ML²/3 for a thin rod about its centre or one end (Serway and Jewett, Table 10.2).

### How do you calculate torque?

Torque is τ = rF sin θ, where r is the distance from the axis to the point where the force acts and θ is the angle between the arm and the force. A 50 N push at 90° on a 0.3 m wrench gives 15 N·m; the same push along the handle (θ = 0°) gives none. One newton-metre is 0.7376 pound-force feet.

### How accurate is the centripetal force, moment of inertia and torque calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “1000 kg car at 20 m/s on a 50 m curve” is checked against Python 3.8 decimal: a = 400/50 = 8, F = 8000, ω = v/r = 0.4.

### Where does the method come from?

OpenStax University Physics Volume 1, §6.3 Centripetal force; §10.5 Calculating moments of inertia; §10.6 Torque; Serway & Jewett, Physics for Scientists and Engineers, Table 10.2 — Moments of inertia of homogeneous rigid objects.

## Sources

- [OpenStax University Physics Volume 1, §6.3 Centripetal force; §10.5 Calculating moments of inertia; §10.6 Torque](https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia)
- Serway & Jewett, Physics for Scientists and Engineers, Table 10.2 — Moments of inertia of homogeneous rigid objects