Question 209: Work, Energy and Power & Circular Motion

Question

The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed of the particle at latitude $30^{\circ}$ is
$$\left[\begin{array}{l}
\sin 30^{\circ}=\cos 60^{\circ}=1 / 2 \
\cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2
\end{array}ight]$$

Accepted Answer

The particle on the equator of the earth has linear speed V.
$$\Rightarrow \mathrm{V}=\mathrm{R} \omega\quad	ext{.... (i)}$$
where R is the radius of the earth.
If the particle is now at $30^{\circ}$ latitude, the particle will move in a circle of smaller radius.
Let the radius of this smaller circle be r .
$$\Rightarrow \mathrm{r}=\mathrm{R} \cos 	heta$$
$	herefore \quad$ The linear velocity of the particle at $30^{\circ}$,
$$\begin{aligned}
V^{\prime} & =r \omega \
& =R \cos 	heta \omega=R \cos 30^{\circ} \omega=\frac{\sqrt{3}}{2} R \omega \quad	ext{.... (ii)}
\end{aligned}$$
Dividing equation (ii) by (i),
$$\begin{aligned}
\frac{V^{\prime}}{V} & =\frac{\frac{\sqrt{3}}{2} R \omega}{R \omega} \
	herefore \quad V^{\prime} & =V \frac{\sqrt{3}}{2}
\end{aligned}$$

Answer

$\frac{\mathrm{V}}{\sqrt{2}}$

Answer

$\frac{\mathrm{V}}{2}$

Answer

$\frac{\sqrt{3}}{2} \mathrm{v}$

Answer

$\mathrm{V}$

Work, Energy and Power & Circular Motion

The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed of the particle at latitude 30^{\circ} is $\left[\begin{array}{l} \sin 30^{\circ}=\cos 60^{\circ}=1 / 2 \\ \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2 \end{array}\right]

Choose the correct answer: