Question

A uniform rod of mass $M$ and length $L$ is free to rotate in $XZ$ plane. A Force $\overrightarrow{F}=(3\hat{i}+2\hat{j}+6\hat{k})N$ is acting on the rod at $(\frac{L}{2},0,0)$ in the situation shown in the figure. The angular acceleration of the rod is : (Take $M=6kg$ and $L=4m$)

• A. $-\frac{3}{2}\hat{j}+\frac{1}{2}\hat{k}$
• B. $-\frac{3}{2}\hat{j}$
• C. $\frac{1}{2}\hat{k}$
• D. $4\hat{j}$

Answer 1

Answer: (B)

$-\frac{3}{2}\hat{j}$

$I=\frac{M{L}^2}{12}=\frac{6\times 4\times 4}{12}=8kg{m}^2$
From $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=[2\hat{i}\times (3\hat{i}+2\hat{j}+6\hat{k}\left)\right]=4\hat{k}-12\hat{j}$
${\tau }_y=I\alpha \Rightarrow -12j=8\alpha \Rightarrow =-\frac{3}{2}\hat{j}rad/s^2$