Question

Consider a body of mass $1.0kg$ at rest at the origin at time $t=0$. A force $\overrightarrow{F}=(\alpha t\hat{i}+\beta \hat{j})$ is applied on the body, where $\alpha =1.0N{s}^{-1}$ and $\beta =1.0N$. The torque acting on the body about the origin at time $t=1.0s$ is $\overrightarrow{\tau }$. Which of the following statements is (are) true?

• A. $\left(\overrightarrow{\tau }\right|=\frac{1}{3}Nm$
• B. The torque $\overrightarrow{\tau }$ is in the direction of the unit vector $+\hat{k}$
• C. The velocity of the body at $t=1s$ is $\overrightarrow{v}=\frac{1}{2}(\hat{i}+2\hat{j})m{s}^{-1}$
• D. The magnitude of displacement of the body at $t=1s$ is $\frac{1}{6}m$

Answer 1

Answer: (A), (C)

The velocity of the body at $t=1s$ is $\overrightarrow{v}=\frac{1}{2}(\hat{i}+2\hat{j})m{s}^{-1}$

Acting force into a body,
$\overrightarrow{F}=\alpha t\hat{i}+\beta \hat{j}$
[At $t=0,v=0,\overrightarrow{r}=\overrightarrow{0}$]

$\alpha =1$, $\beta =1$
$\overrightarrow{F}=t\hat{i}+\hat{j}$
$m\frac{d\overrightarrow{v}}{dt}=t\hat{i}+\hat{j}$
On integrating we get,

$m\overrightarrow{v}=\frac{t^2}{2}\hat{i}+t\hat{j}$ $[m=1kg]$
$\frac{d\overrightarrow{r}}{dt}=\frac{t^2}{2}\hat{i}+t\hat{j}$ $[\overrightarrow{r}=\overrightarrow{0}$ at $t=0]$

Again integrating w.r.t. $t$ we get,
$\overrightarrow{r}=\frac{t^3}{6}\hat{i}+\frac{t^2}{2}\hat{j}$
At $t=1s$, $\overrightarrow{\tau }=(\overrightarrow{r}\times \overrightarrow{F})=$$\left(\frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}\right)\times (\hat{i}+\hat{j})$

$\overrightarrow{\tau }=-\frac{1}{3}\hat{k}$

$\overrightarrow{v}=\frac{t^2}{2}\hat{i}+t\hat{j}$

At,

At $t=1\overrightarrow{s}=\overrightarrow{r_1}-\overrightarrow{r_0}$

$=\left(\frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}\right]-\left[\overrightarrow{0}\right]$

$\overrightarrow{s}=\frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}$
by taking the magnitude,
$\left|\overrightarrow{s}\right|=\sqrt{{\left(\frac{1}{6}\right)}^2+{\left(\frac{1}{2}\right)}^2}\Rightarrow \frac{\sqrt{10}}{6}m$