A particle of mass $m$ moves in a circle of radius $R$ with constant speed $v$. Its angular momentum about the centre is:
A$mvR$
B$\dfrac{mv}{R}$
C$\dfrac{1}{2}mv^2R$
D$mv^2$
Answer & Solution
Correct answer: A. $mvR$
Angular momentum of a particle about a point $\vec L = \vec r \times \vec p$. For circular motion, $\vec r \perp \vec v$, so magnitude $L = mvR$.
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