Question

The position vector of a particle of mass 2 kg is given as function $\overrightarrow{r}=30t\hat{i}+(40t-5t^2)\hat{j}$. The angular momentum of the particle about the origin at t = 4 s is

• A. 4800 $\frac{kg.m^2}{s}$
• B. 3600 $\frac{kg.m^2}{s}$
• C. 1200 $\frac{kg.m^2}{s}$
• D. 2400 $\frac{kg.m^2}{s}$

Answer 1

The correct option is A 4800 $\frac{kg.m^2}{s}$

$\overrightarrow{r}=30t\hat{i}+(40t-5t^2)\hat{j}$
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$ $\Rightarrow \frac{d(30t\hat{i}+(40t-5t^2\left)\hat{j}\right)}{dt}$
$\overrightarrow{v}=30\hat{i}+(40-10t)\hat{j}$
$t=4s$ ;
$\overrightarrow{r}=(120\hat{i}+80\hat{j})m$
​​​​​​$\overrightarrow{v}=\left(30\hat{i}\right)m/s$
So angular momentum =
$\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}=(120\hat{i}+80\hat{j})\times 2\left(30\hat{i}\right)$
$=(-4800\hat{k}$ )$\frac{kg.m^2}{s}$
⇒$\overrightarrow{\left(L\right|}=4800\frac{kg{m}^2}{s}$