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

Answer: 10

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