Question

Point $A$ of rod $AB$ $(l=2\text{m})$ is moving upwards against a wall with velocity $v=2\text{m/s}$. Find angular speed of the rod at an instant when $\theta ={60}^{\circ }$

• A. $1\text{rad/s}$
• B. $2\text{rad/s}$
• C. $3\text{rad/s}$
• D. $4\text{rad/s}$

Answer 1

The correct option is B $2\text{rad/s}$

By the diagram shown here,

Using right-angled triangle $\Delta OAB$
$y=AB\sin \theta =l\sin \theta$

Now differentiating both sides with respect to time we get,

$\frac{dy}{dt}=l\cos \theta \frac{d\theta }{dt}$

we also know that derivative of position with respect to time is velocity$\left(v\right)$ and the rate of change of angular position of a rotating body is angular velocity $\left(\omega \right)$

$v=l\cos \theta .\omega$
Given, $v=2m/s,l=2m,\theta ={60}^{\text{o}}$
$\Rightarrow \left(\cos {60}^{\text{o}}\right)=\frac{1}{2}$

$\Rightarrow \omega =\frac{2}{2\times \frac{1}{2}}=2\text{rad/s}$

Hence option B is the correct answer.