Relation between Torque and Angular Momentum

The angular momentum of a rotating rigid body is, L = I ω. Differentiating the above equation with respect to time,

dL/ dt = I ( dω / dt) = I α

where α = ( dω / dt), angular acceleration of the body.

But torque τ = I α

Therefore, torque τ = dL/dt

Thus the rate of change of angular momentum of a body is equal to the external torque acting upon the body.

The angular momentum of a rotating rigid body is, L = I ω

The torque acting on a rigid body is, t = dL / dt

When no external torque acts on the system, t = dL / dt = 0

(i.e) L = I ω = constant

Total angular momentum of the body = constant

Angular momentum is perpendicular to the plane formed by the pair of position and linear momentum vectors or by the pair of position and velocity vector, depending upon the formula used. Besides, it is also perpendicular to each of operand vectors. However, the vector relation by itself does not tell which side of the plane formed by operands is the direction of torque.

More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:

τ = r × F,

where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by

τ = rFsinθ,

where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,

τ = rF_⊥

where F_⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.