Example 1
75.0 kg, v = 18 m/s, r = 60 m.
Centripetal Force Calculator
Type m [kg], v [m/s] and r [m]. The calculator computes F = m v²/r: 75 kg, 18 m/s and 60 m is 405 N. Same as F = m a at a = 5.4 m/s². Twice v, four times F.
Just a: a = v²/r. In general: F = m·a. v from ω: v = ω·r.
Result
Mass m (kg), Speed v (m/s) and Radius r (m). The result shows up here.
How it works
Centripetal force is F = m v² / r. At 75 kg, 18 m/s and r = 60 m: v² = 324, F = 24300 / 60 = 405 N. The same a = 324 / 60 = 5.4 m/s², so F = 75 × 5.4 = 405 N. That is not a new kind of force, only a role: something must pull inward so the path stays circular.
The form has three fields: m in kilograms, v in m/s, r in meters. The result is in newtons. v is squared, so twice v is four times F at the same r. A comma and a period are the same v: 18,5 and 18.5.
m and r must be positive. r = 0 does not divide. Typed 0 in v gives F = 0: no speed, no turn. The other fields still need a number before 405 N can appear.
Just a lives on centripetal acceleration; then you multiply by m on F = m a. Here the force in one step. v from ω lives on v = ω r.
Physically the role is tire friction, rope tension, or gravity. The calculator does not ask where F comes from. It asks for m, v, and r.
Type 75, 18, and 60, click Calculate, and match 405 N. The header symbol does not hold you on the arc. Treat 405 N as the inward value, not as weight.
How to use
- In the first field enter mass m in kilograms, for example 75.
- In the second enter v in m/s, for example 18. In the third, r in meters, for example 60.
- Click Calculate. The calculator computes m v² / r. 75, 18 and 60 give 405 N, because a = 5.4 m/s².
- r must be greater than zero. Typed 0 in v gives F = 0. All three fields need a number.
- For a alone, open centripetal acceleration. General F = m a is next door.
Formula
F = m · v2 / r
Equivalently F = m · a with a = v2/r. The force points toward the center.
F, m, v, and r for centripetal force
Centripetal force here is F = m v²/r. 75 kg, 18 m/s and r = 60 m is 405 N. a = 5.4 m/s². Twice v, four times F.
- F
- Inward force [N]. 75 × 324 / 60 = 405 N. A role, not a new interaction.
- m
- Mass [kg]. 75 kg. At v = 0 you get F = 0.
- v
- Tangential speed [m/s]. 18 m/s. v is squared.
- r
- Path radius [m]. 60 m. r must be positive.
Real-life examples
Example 2
0.40 kg at 1.5 m with 6 m/s.
Example 3
Child 35.0 kg, v = 7 m/s, r = 5 m.
Example 4
250 kg (with rider), 25 m/s, r = 80 m.
Example 5
Sample 0.02 kg, v = 40 m/s, r = 0.12 m.
Example 6
70.0 kg, 10 m/s, r = 20 m.
Example 7
Car 40000 kg, 30 m/s, r = 500 m.
Example 8
0.80 kg, 15 m/s, r = 8 m.
Example 9
55.0 kg, 11 m/s, r = 22 m.
Example 10
m = 500 kg, v ≈ 7800 m/s, r ≈ 6771000 m.
Frequently asked questions
How much F at 75 kg, 18 m/s and r = 60 m?
Force is 405 N, acceleration a = 5.4 m/s². That is 75 × 324 / 60. A bend, not a straight.
Which units do I type?
Mass m [kg], speed v [m/s], radius r [m]. Result F [N]. Convert 50 km/h to about 13.9 m/s first.
What if v = 0?
Force is 0 N. No speed, no turn. All three fields still need a number.
Where is a = v²/r alone?
On the centripetal-acceleration page. Here the force F = m v²/r, with mass in the form.
Is centripetal force a special kind of force?
It names a role: the inward force for a circular path. Physically friction, tension or gravity, not a fourth fundamental.
Why is v squared?
Because a = v²/r. Doubling v multiplies F by four at the same r and m.
Does a comma in 18.5 work?
Yes. 18.5 and 18,5 are the same v [m/s]. A comma and a period mean the same value.
What if r is very small?
F grows. r must be positive, otherwise the calculator will not divide by the radius.
How does this relate to F = m·a?
It is exactly that. a = v²/r, so F = m v²/r. General F = m a is next door, with no r.
Where do I get v if I know ω?
From the v = ω r card. Then come back with m, v and r. This field already wants metres per second.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.