The linear speed of a particle at the equator | vedclass

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Question

(C) The particle on the equator of the earth has linear speed $V = R \omega$,where $R$ is the radius of the earth and $\omega$ is the angular velocity

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} V$

Vedclass · Answer

(C) The particle on the equator of the earth has linear speed $V = R \omega$,where $R$ is the radius of the earth and $\omega$ is the angular velocity of the earth's rotation.At a latitude $	heta$,the particle moves in a circle of radius $r = R \cos 	heta$.The linear speed $V'$ of the particle at latitude $	heta$ is given by $V' = r \omega = (R \cos 	heta) \omega$.Substituting $	heta = 30^{\circ}$:$V' = R \omega \cos 30^{\circ} = R \omega \left( \frac{\sqrt{3}}{2} ight)$.Since $V = R \omega$,we have $V' = \frac{\sqrt{3}}{2} V$.

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:

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