Reconstruct angular momentum from a torque history

Keep the reference axis fixed

For the selected fixed-axis system, net external torque equals dL/dt. Its signed time integral is angular impulse: J_L=integral tau dt=Lf−Li. Individual external torques can deliver nonzero impulses even when their sum cancels. A torque-time graph gives angular impulse through signed area; the slope of an angular-momentum-time graph gives net torque.

Integrate with the initial condition

Here I=2 kg m², omega0=−1 rad/s, so L0=−2 kg m²/s. The stipulated net torque is A t(2−t)−B from 0 to 2 s, with coefficients in consistent SI units. Integrating gives L(t)=−2+A(t²−t³/3)−Bt. The total angular impulse is 4A/3−2B. Dividing the result by I gives omega only because I is constant.

A second integral describes angle

With theta0=0, theta(t)=−t+A(t³/6−t⁴/24)−Bt²/4. Angular impulse and displacement are different integrals: at 2 s theta=−2+2A/3−B. A body can have zero final angular momentum yet have undergone a nonzero angle and nonzero intermediate torques. If I changed, integrate L first and use omega(t)=L(t)/I(t), rather than treating tau=I alpha with constant I.

Predict the graph before checking

At an instant where torque is negative, L has negative slope, regardless of whether L itself is positive or negative. At a zero of torque, the slope vanishes; this need not mean L=0. Compare a flat momentum segment with a momentum curve crossing zero. Only the first establishes zero net torque throughout that segment.

Worked original case

For A=3 and B=1, impulse is 2 N m s. Starting at L=−2 gives Lf=0 and omega_f=0, but theta(2)=−1 rad. The average net torque is 1 N m; it is not the instantaneous torque at every time.

Investigation

Use A=3,B=1 and then A=3,B=2. Predict signed areas and final momenta. Differentiate your reconstructed L at t=0,1,2 and compare with the torque function. Integrate omega independently and explain why zero net impulse does not imply zero torque at every instant.

Public numerical transfer practice

Try a new practice case

A system starts with angular momentum −3 kg m²/s and receives constant net torque 2 N m for 4 s. What is final signed angular momentum in kg m²/s?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax angular momentum. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.