Conservation of Angular Momentum — Question 7

Question 7

After a tether is cut angular speed falls while angular momentum stays fixed

A particle of mass $m$ is at $(R,0)$ moving at $(0,V)$ when its tether is cut. No external force acts afterward.

1. Find its angular momentum about the original tether point for $t>0$.

2. Find its polar angular speed and explain why it is not constant.

Question 7 - Solution

After a tether is cut angular speed falls while angular momentum stays fixed

(a) Derivation and calculation

After the cut the motion is force-free: $$\vec r(t)=R\hat x+Vt\hat y,\qquad\vec p=mV\hat y.$$ Consequently, $$\boxed{\vec L=\vec r\times\vec p=mRV\hat z}\quad\text{for all }t>0.$$

(b) Result and interpretation

The polar angular rate is $$\dot\theta=\frac{x\dot y-y\dot x}{x^2+y^2} =\boxed{\frac{RV}{R^2+V^2t^2}}.$$ Equivalently, $$mr^2\dot\theta=m(R^2+V^2t^2)\frac{RV}{R^2+V^2t^2}=mRV.$$ The distance from the origin grows, so the angular rate decreases even though angular momentum remains constant.

Physical check

The rate starts at $V/R$ and approaches zero along the straight outgoing trajectory.