Question

A particle of mass ' $m$ ' is moving in time '$t$' on a trajectory given by
$\vec{r}=10 \alpha t^{2}\hat{i}+5 \beta(t-5) \hat{j}$
Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for $t=0$ at time $t=$ seconds.

Answer 1

$\vec{r}=10 \alpha t^{2} \hat{i}+5 \beta(t-5) \hat{j}$
$\vec{v}=20 \alpha \hat{i}+5 \beta \hat{j}$
$\vec{L}=m(\vec{r} \times \vec{v})$
$=m\left[10 \alpha t^{2} \hat{i}+5 \beta(t-5) \hat{j}\right] \times[20 \alpha t \hat{i}+5 \beta \hat{j}]$
$\vec{L}=m\left[50 \alpha \beta t^{2} \hat{k}-100 \alpha \beta\left(t^{2}-5 t\right) \hat{k}\right]$
At $t=0, \vec{L}=\vec{0}$
$50 \alpha \beta t^{2}-100 \alpha \beta\left(t^{2}-5 t\right)=0$
$t-2(t-5)=0$
$t=10\, \sec$