Rotational Motion

JEE Main level, single correct, medium

221 of 280

Question 221 of 280

The time dependence of the position of a particle of mass m=2m=2 is given by r⃗(t)=2ti^−3t2j^\vec{r}\left(t\right)=2t\hat{i}-3{t}^{2}\hat{j} . Its angular momentum, with respect to the origin, at time t=2t=2 is:

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Answer: (C) −48k^-48\hat{k}

r⃗=2ti^−3t2j^\vec{r}=2t \hat{i}-3{t}^{2}\hat{j}
v⃗=2i^−6tj^\vec{v}=2\hat{i}-6t\hat{j}
Momentum
p⃗=mv⃗\vec{p}=m\vec{v}
=4i^−12tj^=4\hat{i}-12t\hat{j}
∴∴ angular momentum L⃗=r⃗×p⃗\vec{L}=\vec{r}\times \vec{p}
=(2ti^−3t2j^)×(4i^−12tj^)=\left(2t\hat{i}-3{t}^{2}\hat{j}\right)\times \left(4\hat{i}-12t\hat{j}\right)
=−24t2k^+12t2k^=-24{t}^{2}\hat{k}+12{t}^{2}\hat{k}
=−12t2k^=-12{t}^{2}\hat{k}
At time
t=2t=2
L⃗=−48k^\vec{L}=-48\hat{k}

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