r=2ti^−3t2j^
v=2i^−6tj^
Momentum p=mv
=4i^−12tj^
∴ angular momentum L=r×p
=(2ti^−3t2j^)×(4i^−12tj^)
=−24t2k^+12t2k^
=−12t2k^
At time t=2
L=−48k^
JEE Main 2019 — Physics Mechanics
The time dependence of the position of a particle of mass m=2 is given by r(t)=2ti^−3t2j^ . Its angular momentum, with respect to the origin, at time t=2 is:
Held on 10 Apr 2019 · Verified 6 Jul 2026.
36k^
48(i^+j^)
−48k^
−34(k^−i^)
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