Question 397: Centre of Mass and Rotational Motion

Question

The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of $\vec{F} = \hat{i} - \hat{j} + \hat{k}$ acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is __________.

Accepted Answer

The torque $\vec{	au}$ acting on the particle with respect to the origin can be calculated using the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$$ \vec{	au} = \vec{r} 	imes \vec{F} $$
Given the position vector $\vec{r} = (1, 1, 1) \, 	ext{m}$ and the force vector $\vec{F} = \hat{i} - \hat{j} + \hat{k}$, we need to calculate the cross product:
$$ \vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -1 & 1 \end{vmatrix} $$
Calculating the determinant, we have:
$$ \vec{	au} = \hat{i} \left(1 \cdot 1 - 1 \cdot (-1)ight) - \hat{j} \left(1 \cdot 1 - 1 \cdot 1ight) + \hat{k} \left(1 \cdot (-1) - 1 \cdot 1ight) $$
This simplifies to:
$$ \vec{	au} = \hat{i}(1 + 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1) $$
$$ \vec{	au} = 2\hat{i} - 0\hat{j} - 2\hat{k} $$
The torque vector is $\vec{	au} = 2\hat{i} - 2\hat{k}$.
To find the magnitude of the torque in the z-direction, we look at the $\hat{k}$ component:
$$ 	au_z = -2 $$
The magnitude of torque in the z-direction is:
$$ |	au_z| = 2 \, 	ext{Nm} $$
Thus, the magnitude of the torque in the z-direction is 2 Newton-meters.

Centre of Mass and Rotational Motion

The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of \vec{F} = \hat{i} - \hat{j} + \hat{k} acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is __________.

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