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The time dependence of the position of a particle of mass m=2 is given by ⃗r(t)=2t̂i-3t²̂j. Its angular momentum about the origin at t=2 s is

Asked in JEE Main 10th April 2nd Shift 2019 · Angular momentum of a particle

(1) -48̂k
(2) -34(̂k-̂i)
(3) 36̂k
(4) 48(̂i+̂j)

Answer: (1) -48̂k

Step-by-step solution

⃗v=(d⃗r)/(dt)=2̂i-6t̂j. ⃗L=m(⃗r×⃗v).

⃗r×⃗v=(2t̂i-3t²̂j)×(2̂i-6t̂j)=(-12t²+6t²)̂k=-6t²̂k.

⃗L=2(-6t²)̂k=-12t²̂k; at t=2, ⃗L=-48̂k.

Why the other options are wrong

(2) Comes from adding spurious ̂i terms from mis-taking the cross product.
(3) Comes from dropping the mass factor and a sign.
(4) Comes from mishandling ̂i×̂j so the result lands in the plane.

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