Question

Two moving particles $P$ and $Q$ are $10\text{m}$ apart at a certain instant. The velocity of $P$ is $8\text{m/s}$ making an angle of ${30}^{\circ }$ with the line joining $P$ and $Q$. The velocity of $Q$ is $6\text{m/s}$ making an angle of ${30}^{\circ }$ with the line joining $P$ and $Q$ as shown in figure. The angular velocity of $P$ with respect to $Q$ is

• A. $0\text{rad/s}$
• B. $0.1\text{rad/s}$
• C. $0.4\text{rad/s}$
• D. $0.7\text{rad/s}$

Answer 1

The correct option is D $0.7\text{rad/s}$

Given that,
Distance between the particle, $r=10\text{m}$
Velocity of particle $P,v_P=8\text{m}$
Velocity of particle $Q,v_Q=6\text{m}$
Let the mass of the particle $P$ is $m$


Brake the velocity into its components
$\Rightarrow v_P=v_P\cos {30}^{\circ }\hat{i}-v_P\sin {30}^{\circ }\hat{j}$
$\Rightarrow v_P=\frac{8\sqrt{3}}{2}i-\frac{8}{2}j$
Similarly,
$\Rightarrow v_Q=v_Q\cos {30}^{\circ }\hat{i}+v_Q\sin {30}^{\circ }\hat{j}$
$\Rightarrow v_Q=\frac{6\sqrt{3}}{2}i+\frac{6}{2}j$

The velocity of $P$ w.r.t $Q$ is
${\overrightarrow{v}}_{PQ}={\overrightarrow{v}}_P-{\overrightarrow{v}}_Q$
$\Rightarrow {\overrightarrow{v}}_{PQ}=(\frac{8\sqrt{3}}{2}i-\frac{8}{2}j)-(\frac{6\sqrt{3}}{2}i+\frac{6}{2}j)$
$\Rightarrow {\overrightarrow{v}}_{PQ}=(\sqrt{3}i-7j)\text{m/s}$
The angular moment$\left(L\right)$ of $P$ w.r.t. $Q$ is
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$ ($\overrightarrow{p}=$Linear momentum)
$\overrightarrow{L}=(-10i)\times m(\sqrt{3}i-7j)=70\text{m}\hat{k}$

The angular moment is also given by
$L=I\omega =m\times r^2\times w$, where $I$ is moment of inertia, $\omega$ angular velocity of $P$ with respect to $Q$
$m\times {10}^2\times \omega =m\times 70$

$\Rightarrow \omega =0.7\text{rad/s}$

Hence, option (d) is correct.