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Worked example 9.1: Angular momentum of a missile

Question: A missile of mass $m=2.3\times 10^4 {\rm kg}$ flies level to the ground at an altitude of $d=10,000 {\rm m}$ with constant speed $v= 210 {\rm m/s}$. What is the magnitude of the missile's angular momentum relative to a point on the ground directly below its flight path?

\epsfysize =2in

Answer: The missile's angular momentum about point $O$ is

L = m v r \sin\theta,

where $\theta$ is the angle subtended between the missile's velocity vector and its position vector relative to $O$. However,

r \sin\theta = d,

where $d$ is the distance of closest approach of the missile to point $O$. Hence,

L = m v d = (2.3\times 10^4)\times 210\times (1\times 10^4)= 4.83\times 10^{10} {\rm kg m^2/s}.

Richard Fitzpatrick 2006-02-02