Question 508: Centre of Mass and Rotational Motion

Question

If $\vec{F}=(5 \hat{i}-10 \hat{j})$ and $\vec{r}=(4 \hat{i}-3 \hat{j})$, then the torque acting on the object will be

Accepted Answer

Given:
$ \vec{F} = 5\hat{i} - 10\hat{j} $
$ \vec{r} = 4\hat{i} - 3\hat{j} $
Step 1:
Torque ($\vec{	au}$) is given by the cross product:
$ \vec{	au} = \vec{r} 	imes \vec{F} $
Step 2:
Write vectors in component form:
$ \vec{r} = 4\hat{i} + (-3)\hat{j} + 0\hat{k} $
$ \vec{F} = 5\hat{i} + (-10)\hat{j} + 0\hat{k} $
Step 3:
Use the determinant method for cross product:
$ \vec{	au} =  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -3 & 0 \ 5 & -10 & 0 \end{vmatrix} $
Step 4:
Expand the determinant:
$ \vec{	au} = \hat{i} \begin{vmatrix} -3 & 0 \ -10 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 0 \ 5 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & -3 \ 5 & -10 \end{vmatrix} $
Calculate each minor:
$\displaystyle\begin{vmatrix} -3 & 0 \ -10 & 0 \end{vmatrix}= (-3)(0) - (0)(-10) = 0$
$\displaystyle\begin{vmatrix} 4 & 0 \ 5 & 0 \end{vmatrix}= (4)(0) - (0)(5) = 0$
$\displaystyle\begin{vmatrix} 4 & -3 \ 5 & -10 \end{vmatrix} = (4)(-10) - (-3)(5) = -40 + 15 = -25$
So,
$ \vec{	au} = 0\hat{i} - 0\hat{j} -25\hat{k} $
$ \vec{	au} = -25\hat{k} $
Final Answer:
$ \boxed{-25\hat{k}} $
So, the correct option is
Option D
.

Answer

$\hat{\mathrm{i}}-2 \hat{\mathrm{j}}$

Answer

$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}$

Answer

$25 \hat{\mathrm{k}}$

Answer

$-25 \hat{\mathrm{k}}$

Centre of Mass and Rotational Motion

If \vec{F}=(5 \hat{i}-10 \hat{j}) and \vec{r}=(4 \hat{i}-3 \hat{j}), then the torque acting on the object will be

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