Rotation Ques 35
- The time dependence of the position of a particle of mass $m=2$ is given by $\mathbf{r}(t)=2 t \hat{\mathbf{i}}-3 t^{2} \hat{\mathbf{j}}$. Its angular momentum, with respect to the origin, at time $t=2$ is
(a) $36 \hat{\mathbf{k}}$
(b) $-34(\hat{\mathbf{k}}-\hat{\mathbf{i}})$
(c) $-48 \hat{\mathbf{k}}$
(d) $48(\hat{\mathbf{i}}+\hat{\mathbf{j}})$
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Answer:
Correct Answer: 35.(c)
Solution:
- Position of particle is, $\mathbf{r}=2 t \hat{\mathbf{i}}-3 t^{2} \hat{\mathbf{j}}$
where, $t$ is instantaneous time.
Velocity of particle is
$$ \mathbf{v}=\frac{d \mathbf{r}}{d t}=2 \hat{\mathbf{i}}-6 \hat{t} \mathbf{j} $$
Now, angular momentum of particle with respect to origin is given by
$$ \begin{gathered} \mathbf{L}=m(\mathbf{r} \times \mathbf{v}) \\ =m{\left(2 t \hat{\mathbf{i}}-3 t^{2} \hat{\mathbf{j}}\right) \times(2 \hat{\mathbf{i}}-6 t \hat{\mathbf{j}}) } \end{gathered} $$
As, $\quad \hat{\mathbf{i}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}} \times \hat{\mathbf{j}}=0$
$\Rightarrow \quad \mathbf{L}=m\left(-12 t^{2} \hat{\mathbf{k}}+6 t^{2} \hat{\mathbf{k}}\right)$
As, $\quad \hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}}$ and $\hat{\mathbf{j}} \times \hat{\mathbf{i}}=-\hat{\mathbf{k}}$
$\Rightarrow \quad \mathbf{L}=-m\left(6 t^{2}\right) \hat{\mathbf{k}}$
So, angular momentum of particle of mass $2 kg$ at time $t=2 s$ is
$$ \mathbf{L}=\left(-2 \times 6 \times 2^{2}\right) \hat{\mathbf{k}}=-48 \hat{\mathbf{k}} $$
BETA
