Question

A rod of length $1\text{m}$ rotates in the $xy$ plane about the fixed point $\text{O}$ in the anticlockwise sense, as shown in figure, with angular velocity $\omega =a+bt$ where $a=10{\text{rad s}}^{-1}$ and $b=5{\text{rad s}}^{-2}$. The velocity and acceleration of the point $A$ at $t=0$ is

• A. $+10\hat{j}{\text{ms}}^{-1}\text{and}5\hat{j}{\text{ms}}^{-2}$
• B. $+10\hat{j}{\text{ms}}^{-1}\text{and}(-100\hat{i}+5\hat{j}){\text{ms}}^{-2}$
• C. $-10\hat{j}{\text{ms}}^{-1}\text{and}(100\hat{i}+5\hat{j}){\text{ms}}^{-2}$
• D. $+10\hat{j}{\text{ms}}^{-1}\text{and}-5\hat{j}{\text{ms}}^{-2}$

Answer 1

The correct option is B $+10\hat{j}{\text{ms}}^{-1}\text{and}(-100\hat{i}+5\hat{j}){\text{ms}}^{-2}$

Given $\omega =10+5t$
$\Rightarrow \alpha =\frac{d\omega }{dt}=5$
At $t=0,\omega =10\text{rad/s}$
and $\alpha =5{\text{rad/s}}^2$
and $r=OA=1\text{m}$
Velocity of point $\text{A}$ at $t=0$
$v=\left(r\omega \right)\hat{j}=\left(10\hat{j}\right)\text{m/s}$

Net acceleration of point $\text{A}$ at $t=0$ = tangential acceleration + centripetal acceleration
$=\left(r\alpha \right)\hat{j}-\left(r{\omega }^2\right)\hat{i}$
$=(-100\hat{i}+5\hat{j}){\text{m/s}}^2$