Question

The position vector $\vec{r}$ of a particle of mass $m$ is given by the following equation $\vec{r}(t)=\alpha t^{3} \hat{i}+\beta t^{2} \hat{j}$ where $\alpha=\frac{10}{3} \,ms ^{-3}, \beta=5\, ms ^{-2}$ and $m =0.1 \,kg$. At $t =1\, s$, which of the following statement (s) is (are) true about the particle?

• A. The velocity $\overrightarrow{ v }$ is given by $\overrightarrow{ v }=(10 \hat{ i }+10 \hat{ j }) ms ^{-1}$
• B. The angular momentum $\overrightarrow{ L }$ with respect to the origin is given by $\overrightarrow{ L }=-\left(\frac{5}{3}\right) \hat{ k } N ms$
• C. The force $\overrightarrow{ F }$ is given by $\overrightarrow{ F }=(\hat{ i }+2 \hat{ j }) N$
• D. The torque $\vec{\tau}$ with respect to the origin is given by $\vec{\tau}=-\left(\frac{20}{3}\right) \hat{k} Nm$

Answer 1

The Correct Option is D

$\vec{r}(t)=\alpha t^{3} \hat{i}+\beta t^{2} \hat{j}$
$\alpha=\frac{10}{3} ms ^{-3}, \beta=5\, ms ^{-2}$ and $m =0.1\, kg . t =1\, s$, $\overrightarrow{ v }=3 \alpha t ^{2} \hat{ i }+2 \beta t \hat{ j }$
(A) $(\overrightarrow{ v })_{ r =1}=3 \alpha \hat{ i }+2 \beta \hat{ j }$
$=3 \times \frac{10}{3} \hat{ i }+2 \times 5 \hat{ j }$
$=10 \hat{ i }+10 \hat{ j }$
(B) $\vec{L} =\vec{r} \times \overrightarrow{ P }= m (\overrightarrow{ r } \times \overrightarrow{ v })$
$=0.1\left[\left(\alpha t ^{3} \hat{i}+\beta t ^{2} \hat{ j }\right) \times(10 \hat{ i }+10 \hat{ j })\right] $
$=0.1\left[\left(10 \alpha t ^{3}(\hat{ k })-10 \beta t ^{2}(\hat{ k })\right]\right.$
$=0.1\left[10 \times \frac{10}{3} \times 1 \hat{ k }-10 \times 5(1)^{2} \hat{ k }\right] $
$=0.1\left[\frac{100}{3}-50\right] \hat{ k } $
$=-\frac{5}{3}(\hat{k})$
(C) $\overrightarrow{ F } = m \overrightarrow{ a }= m [6 \alpha t \hat{ i }+2 \beta \hat{ j }] $
$ 0.1\left[6 \times \frac{10}{3} \times 1 \hat{ i }+2 \times 5 \hat{ j }\right] $
$=[2 \hat{ i }+\hat{ j }] $
(D) $ \vec{\tau} =\overrightarrow{ r } \times \overrightarrow{ F } $
$=\left(\alpha t ^{3} \hat{ i }+\beta t ^{2} \hat{ j }\right) \times[2 \hat{ i }+\hat{ j }] $
$=\left(\frac{10}{3} \hat{ i }+5 \hat{ j }\right) \times[2 \hat{ i }+\hat{ j }] $
$=\frac{10}{3} \hat{ k }-10 \hat{ k } $
$=\frac{-20}{3} \hat{ k } $