Question

A rigid body is spinning about a fixed point $(3,-2,-1)$ with angular velocity of $4$ rad/s, the axis of rotation being in the direction of $(1,2,-2)$, then the velocity of the particle at the point $(4,1,1)$ is?

• A. $\frac{4}{3}(1,-4,10)$
• B. $\frac{4}{3}(4,-10,1)$
• C. $\frac{4}{3}(10,-4,1)$
• D. $\frac{4}{3}(10,4,1)$

Answer 1

The correct option is C $\frac{4}{3}(10,-4,1)$

Let $\omega$ be the angular velocity and $\overrightarrow{r}$ be the position vector $\overrightarrow{AP}=\overrightarrow{r}$, from figure.

Given the direction coordinates, $(1,2,-2)$.

$\therefore$ the angular velocity,

$\omega =magnitude\times direction$

$\omega =\frac{4(\hat{i}+2\hat{j}-2\hat{k})}{\sqrt{1+4+4}}$

$\omega =\frac{4}{3}(\hat{i}+2\hat{j}-2\hat{k})$ ----(i)

Position Vector, $\overrightarrow{AP}=\overrightarrow{r}=\overrightarrow{OP}-\overrightarrow{OA}$

$=(4\hat{i}+\hat{j}+\hat{k})-(3\hat{i}-2\hat{j}-\hat{k})$

$=(\hat{i}+3\hat{j}+2\hat{k})$ ----(ii)

$Velocity,\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$

$=\frac{4}{3}(\hat{i}+2\hat{j}-2\hat{k})\times (\hat{i}+3\hat{j}+2\hat{k})$

$\therefore \overrightarrow{v}=\frac{4}{3}(10\hat{i}-4\hat{j}+\hat{k})$

This is the angular Velocity.