The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} – 6\hat{k}$ at $(2, 0, -3)$, about the point $(2, -2, -2)$ is given by
(2018)
1. $-7\hat{i} - 8\hat{j} - 4\hat{k}$
2. $-4\hat{i} - \hat{j} - 8\hat{k}$
3. $-8\hat{i} - 4\hat{j} - 7\hat{k}$
4. $-7\hat{i} - 4\hat{j} - 8\hat{k}$
View Answer
The relative position vector is $\vec{r} = (2-2)\hat{i} + (0 - (-2))\hat{j} + (-3 - (-2))\hat{k} = 2\hat{j} - \hat{k}$. Torque is $\vec{\tau} = \vec{r} \times \vec{F} = (2\hat{j} - \hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k}) = -7\hat{i} - 4\hat{j} - 8\hat{k}$.
The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its midpoint and perpendicular to its length is $I_{0}$. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is
(2011 Mains)
1. $I_{0} + ML^{2}/2$
2. $I_{0} + ML^{2}/4$
3. $I_{0} + 2ML^{2}$
4. $I_{0} + ML^{2}$
View Answer
Using the parallel axis theorem, $I = I_{cm} + Md^{2}$. Here, the center of mass moment of inertia is $I_{cm} = I_{0}$ and the distance to the parallel axis is $d = \frac{L}{2}$. Thus, $I = I_{0} + M(\frac{L}{2})^{2} = I_{0} + \frac{ML^{2}}{4}$.