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 α=1.0N/s and β=1.0N. The torque acting on the body about the origin at time t=1.0s is τ. Which of the following statements is (are) true?
A. $\left|\overrightarrow{\tau }\right|={\displaystyle \frac{1}{3}}N.m$
B. The torque $\overrightarrow{\tau }$is in the direction of unit vector $+\hat{k}$
C. Velocity of the body at t= 1sec is $\overrightarrow{v}={\displaystyle \frac{1}{2}}(\hat{i}+2\hat{j})$m/s
D. The magnitude of displacement of the body at t= 1s is ${\displaystyle \frac{1}{6}}m$

Answer 1

Hint: From the vector form of 2nd law of Newton,$\overrightarrow{F}=m\overrightarrow{a}=m{\displaystyle \frac{d\overrightarrow{v}}{dt}}$ and we know,$\overrightarrow{v}={\displaystyle \frac{d\overrightarrow{r}}{dt}}$
From the definition of torque,$\overrightarrow{\tau }=(\overrightarrow{r}\times \overrightarrow{F})$
Where, $\overrightarrow{\tau }=$torque acting on the body
$\overrightarrow{F}=$Force acting on the body
$\overrightarrow{r}=$Displacement of the body
$\overrightarrow{v}=$Velocity of the body
$m=$Mass of the body
Using these two equations we will come to the solution of the above problem.

Complete step by step answer:
Mass of the body, $m=1kg$
At $t=0s,\overrightarrow{v}=0,\overrightarrow{r}=0$
Also,$\overrightarrow{F}=\alpha t\hat{i}+\beta \hat{j}$, $\alpha =1N/s,\beta =1N$
So, $\overrightarrow{F}=t\hat{i}+\hat{j}$
From Newton’s second law,$\overrightarrow{F}=m\overrightarrow{a}=m{\displaystyle \frac{d\overrightarrow{v}}{dt}}$
Now, $m{\displaystyle \frac{d\overrightarrow{v}}{dt}}=t\hat{i}+\hat{j}$$$$$
Or, $md\overrightarrow{v}=(t\hat{i}+\hat{j})dt$
Integrating both sides,$m\underset{v=0}{\overset{v=\overrightarrow{v}}{\int }}d\overrightarrow{v}=\underset{t=0}{\overset{t=t}{\int }}(t\hat{i}+\hat{j})dt$
Or, $\overrightarrow{v}={\displaystyle \frac{t^2}{2}}\hat{i}+t\hat{j}[\because m=1kg]$
Or,${\displaystyle \frac{d\overrightarrow{r}}{dt}}={\displaystyle \frac{t^2}{2}}\hat{i}+t\hat{j}$
Or,$d\overrightarrow{r}=({\displaystyle \frac{t^2}{2}}\hat{i}+t\hat{j})dt$
Integrating both sides, $\underset{r=0}{\overset{r=\overrightarrow{r}}{\int }}d\overrightarrow{r}=\underset{t=0}{\overset{t=t}{\int }}\left({\displaystyle \frac{t^2}{2}}\hat{i}+t\hat{j}\right)dt$
Or, $\overrightarrow{r}={\displaystyle \frac{t^3}{6}}\hat{i}+{\displaystyle \frac{t^2}{2}}\hat{j}$
At t= 1sec,
$\overrightarrow{r}={\displaystyle \frac{1^3}{6}}\hat{i}+{\displaystyle \frac{1^2}{2}}\hat{j}={\displaystyle \frac{1}{6}}\hat{i}+{\displaystyle \frac{1}{2}}\hat{j}$
$\overrightarrow{v}={\displaystyle \frac{1^2}{2}}\hat{i}+1\hat{j}={\displaystyle \frac{1}{2}}\hat{i}+1\hat{j}={\displaystyle \frac{1}{2}}(\hat{i}+2\hat{j})$………………………….(1)
$\overrightarrow{F}=\hat{i}+\hat{j}$
$$\begin{gathered} \overrightarrow{\tau }=(\overrightarrow{r}\times \overrightarrow{F}) \\ =({\displaystyle \frac{1}{6}}\hat{i}+{\displaystyle \frac{1}{2}}\hat{j})\times (\hat{i}+\hat{j}) \\ =({\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{2}})\hat{k} \\ =-{\displaystyle \frac{1}{3}}\hat{k}................................(2) \end{gathered}$$
$\left|\overrightarrow{r}\right|=\sqrt{{({\displaystyle \frac{1}{6}})}^2+{({\displaystyle \frac{1}{2}})}^2}={\displaystyle \frac{\sqrt{10}}{6}}$……………………………………(3)
Now, from (2) it is clear that at t= 1sec, $|\overrightarrow{\tau }|={\displaystyle \frac{1}{3}}$.
So, option (A) is correct.
Direction of $\overrightarrow{\tau }$is towards unit vector $-\hat{k}$from (2)
So, option (B) is incorrect.
From (1), Velocity of the body at t= 1sec is $\overrightarrow{v}={\displaystyle \frac{1}{2}}(\hat{i}+2\hat{j})$m/s
So, option (C) is correct.
From (3), the magnitude of displacement of the body at t= 1s is ${\displaystyle \frac{\sqrt{10}}{6}}m$
So, option (D) is incorrect.

So, the correct answers are “Options A and C”.

Note:
It is to be noted that, $\overrightarrow{\tau }=(\overrightarrow{r}\times \overrightarrow{F})$. $(\overrightarrow{r}\times \overrightarrow{F})\ne (\overrightarrow{F}\times \overrightarrow{r})$. So, $\overrightarrow{\tau }\ne (\overrightarrow{F}\times \overrightarrow{r})$.
If the body is not at rest initially and let have a speed of $\overrightarrow{v_1}$, then limits of the integration will change, like,$m\underset{v=\overrightarrow{v_1}}{\overset{v=\overrightarrow{v}}{\int }}d\overrightarrow{v}=\underset{t=0}{\overset{t=t}{\int }}(t\hat{i}+\hat{j})dt$.