Example 1
ω = 10 rad/s, r = 0.35 m.
Type ω [rad/s] and r [m]. The calculator computes v = ω r: 10 rad/s and 0.35 m is 3.5 m/s (12.6 km/h), and 0.8 rad/s at 6 m is 4.8 m/s. On the axis r = 0 the speed vanishes.
Where ω comes from: ω = 2π/T. Then a: a = v²/r. Force: F = m·v²/r.
Angular velocity ω (rad/s) and Radius r (m). The result shows up here.
Rim speed is angular velocity times radius: v = ω r. At 10 rad/s and r = 0.35 m you get 3.5 m/s, or 12.6 km/h. At 0.8 rad/s and r = 6 m you get 4.8 m/s. At 300 rad/s and r = 0.1 m you get 30 m/s. Farther from the axis means larger v at the same ω.
The form has two fields: ω in radians per second and r in meters. The result is in m/s. Degrees per second do not belong here: 360°/s is 2π rad/s, about 6.28, not 360. A comma and a period are the same r, so 0,35 and 0.35 are the same radius.
In rigid rotation every point shares one ω. They differ in r, so they differ in v. On the axis r = 0, so v = 0: the point still turns in angle but covers no arc. Typed 0 in ω also gives v = 0, because nothing is spinning. Both fields need a number before 3.5 m/s can appear.
ω from a period is on the neighbouring page: ω = 2π/T. One turn in 8 s is about 0.785 rad/s; you bring that here with r. From v you go on to a = v²/r. From 3.5 m/s and 0.35 m that is 35 m/s².
Centripetal force F = m v²/r takes the same v. This calculator stops at m/s, not newtons and not m/s².
Type 10 and 0.35, click Calculate, and match 3.5 m/s. The header symbol does not convert radians to degrees. Treat the result as tangential speed at that r, not as angular velocity.
v = ω · r
ω in rad/s, r in meters, v in m/s. In rigid rotation, points at different r share ω.
Rigid rotation: v = ω·r. 10 rad/s and 0.35 m is 3.5 m/s (12.6 km/h). 0.8 rad/s at r = 6 m is 4.8 m/s. On the axis r = 0, v = 0.
ω = 10 rad/s, r = 0.35 m.
ω = 0.8 rad/s, r = 6 m.
ω = 300 rad/s, r = 0.10 m.
ω = 314 rad/s (~3000 rpm), r = 0.10 m.
ω = 0.05 rad/s, r = 25 m.
ω = 126 rad/s, r = 0.2 m.
ω = 80 rad/s, r = 0.3 m.
ω ≈ 3.5 rad/s, r = 0.15 m (rim).
ω = 2 rad/s, r = 0.5 m.
ω ≈ 0.00115 rad/s, r ≈ 6771000 m.
Speed is 3.5 m/s, or 12.6 km/h. That is the product 10 × 0.35. Rim of the wheel, not the hub.
Angular speed ω [rad/s], radius r [m]. Result v [m/s]. Convert degrees per second to radians first.
Speed is 0 m/s. You are standing still, so the product vanishes. Both fields still need a number.
From the angular-velocity card: ω = 2π/T. This field already wants radians per second, not the period.
Because r = 0. Same rotation, zero radius, zero path along the circle. v = ω r dies on the axis.
You get 4.8 m/s. A slower spin, a longer arm, similar speed order to 10 rad/s at 0.35 m.
Yes. 0.35 and 0,35 are the same r [m]. A comma and a period mean the same value.
On the a = v²/r page. Type this v and the same r. Here you only compute v = ω r.
No. The field wants radians per second. 360°/s is 2π rad/s, not 360 in field ω.
Speed is 30 m/s. A fast rotor, a small radius, still highway order on the rim.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.