Question 130: Centre of Mass and Rotational Motion

Question

The torque due to the force $(2 \hat{i}+\hat{j}+2 \hat{k})$ about the origin, acting on a particle whose position vector is $(\hat{i}+\hat{j}+\hat{k})$, would be

Accepted Answer

First, torque ($\vec{	au}$) is found by taking the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$):
$\vec{	au} = \vec{r} 	imes \vec{F}$
Here, $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$.
We use a determinant to find the cross product:

$$ \vec{	au} = \left|  \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 1 & 2 \ \end{array} ight| $$
Expanding the determinant:
$\vec{	au} = \hat{i}[(1)(2)-(1)(1)] -\hat{j}[(1)(2)-(1)(2)] + \hat{k}[(1)(1)-(1)(2)]$
This gives: $\vec{	au} = \hat{i}(2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - 2)$
So, $\vec{	au} = \hat{i} - 0\hat{j} - \hat{k}$

Answer

$\hat{j}+\hat{k}$

Answer

$\hat{i}-\hat{k}$

Answer

$\hat{i}-\hat{j}+\hat{k}$

Answer

$\hat{i}+\hat{k}$

Centre of Mass and Rotational Motion

The torque due to the force (2 \hat{i}+\hat{j}+2 \hat{k}) about the origin, acting on a particle whose position vector is (\hat{i}+\hat{j}+\hat{k}), would be

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