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A particle of mass $2 \mathrm{~kg}$ is moving such that at time $t$, its position, in meter, is given by $\vec{r}(t)=5 \hat{i}-2 t^2 \hat{j}$. The angular momentum of the particle at $t=2 s$ about the origin in $\mathrm{kg} \mathrm{m}^{-2} \mathrm{~s}^{-1}$ is :
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Verified Answer
The correct answer is:
$-80 \hat{k}$
$-80 \hat{k}$
Angular momentum $\mathrm{L}=\mathrm{m}(\mathrm{v} \times \mathrm{r})$
$$
\begin{aligned}
& =2 \mathrm{~kg}\left(\frac{\mathrm{dr}}{\mathrm{dt}} \times \mathrm{r}\right)=2 \mathrm{~kg}\left(4 \mathrm{t} \times \times 5 \mathrm{i}-2 \mathrm{t}^2 \hat{\mathrm{j}}\right) \\
& =2 \mathrm{~kg}(-20 \mathrm{t} \hat{\mathrm{k}})=2 \mathrm{~kg} \times-20 \times 2 \mathrm{~m}^{-2} \mathrm{~s}^{-1} \hat{\mathrm{k}} \\
& =-80 \hat{\mathrm{k}}
\end{aligned}
$$
$$
\begin{aligned}
& =2 \mathrm{~kg}\left(\frac{\mathrm{dr}}{\mathrm{dt}} \times \mathrm{r}\right)=2 \mathrm{~kg}\left(4 \mathrm{t} \times \times 5 \mathrm{i}-2 \mathrm{t}^2 \hat{\mathrm{j}}\right) \\
& =2 \mathrm{~kg}(-20 \mathrm{t} \hat{\mathrm{k}})=2 \mathrm{~kg} \times-20 \times 2 \mathrm{~m}^{-2} \mathrm{~s}^{-1} \hat{\mathrm{k}} \\
& =-80 \hat{\mathrm{k}}
\end{aligned}
$$
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