Question 67: Work, Energy and Power & Circular Motion

Question

As shown in the figure, a particle is moving with constant speed $$\pi ~\mathrm{m} / \mathrm{s}$$. Considering its motion from $$\mathrm{A}$$ to $$\mathrm{B}$$, the magnitude of the average velocity is :

Accepted Answer

The given situation is shown below.
Given, $v=\pi \mathrm{m} / \mathrm{s}$
or
$$
\begin{aligned}
R \omega & =\pi \\
\omega & =\frac{\pi}{R} \mathrm{rad} / \mathrm{s} \quad[\because v=\omega R]
\end{aligned}
$$
Angular displacement, $	heta=120^{\circ}$ or $\frac{2 \pi}{3}$
Using $	heta=\omega t$,
$$
t=\frac{	heta}{\omega}=\frac{2 \pi / 3}{\pi / R}=\frac{2 R}{3}
$$
Linear displacement,
$$
\begin{aligned}
d & =2 R \sin (	heta / 2) \\
d & =2 R \sin \left(\frac{120^{\circ}}{2}ight)=2 R \sin 60^{\circ} \\
& =2 R \frac{\sqrt{3}}{2} \Rightarrow R \sqrt{3}
\end{aligned}
$$
Average velocity $=\frac{d}{t}=\frac{R \sqrt{3}}{2 R / 3}=\frac{3 \sqrt{3}}{2}=1.5 \sqrt{3}\mathrm{~m} / \mathrm{s}$

Answer

$$\pi ~\mathrm{m} / \mathrm{s}$$

Answer

$$1.5 \sqrt{3} \mathrm{~m} / \mathrm{s}$$

Answer

$$\sqrt{3} \mathrm{~m} / \mathrm{s}$$

Answer

$$2 \sqrt{3} \mathrm{~m} / \mathrm{s}$$

Work, Energy and Power & Circular Motion

As shown in the figure, a particle is moving with constant speed $\pi ~\mathrm{m} / \mathrm{s}. Considering its motion from \mathrm{A} to \mathrm{B}$, the magnitude of the average velocity is :

Choose the correct answer: