Torque for Angular Acceleration Calculator
Find the torque needed to angularly accelerate a rotating body (τ = I·α), plus the equivalent force at a given radius.
How it works
τ = I · α · Equivalent tangential force F = τ ÷ r
Newton’s second law for rotation is τ = I·α: torque equals the moment of inertia times the angular acceleration, exactly mirroring F = m·a. A larger moment of inertia — mass spread further from the axis — demands more torque to reach the same angular acceleration. If the torque comes from a force applied tangentially at radius r, then τ = F·r, so the force needed is τ divided by r.
Worked example
To angularly accelerate a wheel of I = 0.5 kg·m² at α = 4 rad/s², the torque is τ = 0.5 × 4 = 2 N·m. If you push tangentially at the rim, radius 0.3 m, the force required is F = 2 / 0.3 ≈ 6.67 N.
Sources: Torque — Wikipedia
Frequently asked questions
How is τ = I·α like F = m·a?
They are the same law in rotational form: torque replaces force, moment of inertia replaces mass, and angular acceleration replaces linear acceleration. Both say the "push" equals inertia times the resulting acceleration.
Why does a larger radius reduce the force needed?
Torque is force times lever arm (τ = F·r). Applying the force further from the axis gives it more leverage, so a smaller force produces the same torque — which is why door handles sit far from the hinge.
What if the angular acceleration is negative?
A negative α means the body is slowing down, so the torque comes out negative — it opposes the rotation (a braking torque). The magnitude still equals I times the rate of change of angular velocity.
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This calculator is for educational purposes. Double-check important results before acting on them.