prove that angular momentum is equal to linear momentum
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Relation of Angular to Linear Momentum
Recall (§B.3) that the momentum of a mass $ m$ traveling with velocity $ v$ in a straight line is given by
$\displaystyle p = m v,
$
while the angular momentum of a point-mass $ m$ rotating along a circle of radius $ R$ at $ \omega $ rad/s is given by
$\displaystyle L \eqsp I\omega,
$
where $ I=mR^2$ . The tangential speed of the mass along the circle of radius $ R$ is given by
$\displaystyle v \eqsp R\omega.
$
Expressing the angular momentum $ I$ in terms of $ v$ gives
$\displaystyle L \isdefs I\omega \eqsp I\frac{v}{R} \isdefs mR^2\frac{v}{R} \eqsp Rmv \eqsp Rp. \protect$ (B.18)
Thus, the angular momentum $ L$ is $ R$ times the linear momentum $ p=mv$ .
Linear momentum can be viewed as a renormalized special case of angular momentum in which the radius of rotation goes to infinity.
Recall (§B.3) that the momentum of a mass $ m$ traveling with velocity $ v$ in a straight line is given by
$\displaystyle p = m v,
$
while the angular momentum of a point-mass $ m$ rotating along a circle of radius $ R$ at $ \omega $ rad/s is given by
$\displaystyle L \eqsp I\omega,
$
where $ I=mR^2$ . The tangential speed of the mass along the circle of radius $ R$ is given by
$\displaystyle v \eqsp R\omega.
$
Expressing the angular momentum $ I$ in terms of $ v$ gives
$\displaystyle L \isdefs I\omega \eqsp I\frac{v}{R} \isdefs mR^2\frac{v}{R} \eqsp Rmv \eqsp Rp. \protect$ (B.18)
Thus, the angular momentum $ L$ is $ R$ times the linear momentum $ p=mv$ .
Linear momentum can be viewed as a renormalized special case of angular momentum in which the radius of rotation goes to infinity.
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ᴍɪss ᴍᴀsᴛᴇʀᴍɪɴᴅ~❤️
In Newtonian mechanics, linear momentum, translational momentum, or simply momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
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