Skip to content
doubtly
Sign up

1 answer

Log in to upvote 0 Log in to downvote

Best answer

Answer: The angular momentum of a body (or system) remains constant if the resultant external torque acting on it is zero.

Proof: Angular momentum of a particle about a point is \(\vec L = \vec r \times \vec p\). Differentiating with respect to time,

\[\frac{d\vec L}{dt} = \frac{d\vec r}{dt}\times \vec p + \vec r \times \frac{d\vec p}{dt} = (\vec v \times m\vec v) + \vec r \times \vec F\]

Since \(\vec v \times \vec v = 0\) and \(\vec r \times \vec F = \vec\tau\),

\[\frac{d\vec L}{dt} = \vec\tau\]

If the external torque \(\vec\tau = 0\), then \(\dfrac{d\vec L}{dt} = 0\), i.e. \(\vec L = \text{constant}\). For a rigid body rotating about a fixed axis, \(L = I\omega\), so \(I_1\omega_1 = I_2\omega_2\). (Example: a spinning ballet dancer or diver pulls in the arms, reducing \(I\), and spins faster.)

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
Was this answer helpful?
Sign in to comment

Your answer

Log in or create a free account to answer this question.