Angular Momentum Of A Particle

Question

A particle of mass 0.5 kg travels with a constant speed of 4 m/s along a straight line whose perpendicular distance from a fixed point O is 3 m. What is the magnitude of its angular momentum about O?

RankGuru · Accepted Answer

The angular momentum of a single particle about a point is $L = m v r_\perp$, where $r_\perp$ is the perpendicular distance from the point to the line of motion. A key conceptual point is that even straight-line motion carries nonzero angular momentum about a point lying off its path. Substituting the given data, $L = (0.5)(4)(3) = 6	ext{ kg m}^2/	ext{s}$. The value 2 kg m$^2$/s uses only $mv$ and forgets the perpendicular distance. The value 12 kg m$^2$/s drops the mass factor of 0.5 and effectively doubles the result. The value 24 kg m$^2$/s omits the mass entirely and mishandles the remaining product. This is the NCERT definition $\vec{L} = \vec{r} 	imes \vec{p}$, here reduced to its magnitude because the line keeps a fixed perpendicular distance of 3 m. As a dimensional check, kilograms times metres per second times metres gives kg m$^2$/s, the correct unit of angular momentum, and the modest input values yield a sensible 6 kg m$^2$/s.

Angular Momentum Of A Particle

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