Question

The position vector $\overrightarrow{r}$ of a particle of mass $0.1\text{kg}$ is given by the equation $\overrightarrow{r}\left(t\right)=(\frac{10}{3}{t}^3\hat{i}+5t^2\hat{j})\text{m}$. At $t=1\text{s}$, the torque experienced by the particle due to force with respect to the origin is given by

• A. $\frac{20}{3}\hat{k}\text{N-m}$
• B. $\frac{20}{3}\hat{j}\text{N-m}$
• C. $-\frac{20}{3}\hat{k}\text{N-m}$
• D. $-\frac{20}{3}\hat{j}\text{N-m}$

Answer 1

The correct option is C $-\frac{20}{3}\hat{k}\text{N-m}$

Given: $\overrightarrow{r}=(\frac{10}{3}{t}^3\hat{i}+5t^2\hat{j})\text{m}$

and,

$\overrightarrow{a}=\frac{d^2\overrightarrow{r}}{d{t}^2}=(20t\hat{i}+10\hat{j}){\text{m/s}}^2$

From Newton's second law , we can write that,

$\overrightarrow{F}=m\overrightarrow{a}=0.1\times (20t\hat{i}+10\hat{j})=(2t\hat{i}+\hat{j})\text{N}$

At $t=1\text{s}$,

$\overrightarrow{r}=\frac{10}{3}\times 1^3\hat{i}+5\times 1^2\hat{j}=(\frac{10}{3}\hat{i}+5\hat{j})\text{m}$

$\overrightarrow{F}=(2\times 1)\hat{i}+\hat{j}=(2\hat{i}+\hat{j})\text{N}$

Now, Torque due to this force with respect to origin is

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{10}{3}& 5& 0\\ 2& 1& 0\end{array}\right|$

$\Rightarrow \overrightarrow{\tau }=\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(\frac{10}{3}-10)\Rightarrow \overrightarrow{\tau }=(-\frac{20}{3}\hat{k})\text{N-m}$

Hence, option (c) is the correct answer.

Note:- For a rigid body rotating about a fixed axis and being acted upon by a force $\text{F}$. Torque of $\text{F}$ about $O$(point on the axis of rotation) is given by $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$