Question

Consider a particle of mass m suspended vertically by a string at the equator. Let R and M denote the radius and the mass of the earth respectively. If $\omega$ is the angular velocity of earth's rotation about its own axis, the tension in the string is equal to

• A. $\frac{GmM}{2R^2}$
• B. $\frac{GmM}{R^2}$
• C. $\frac{GmM}{R^2}-m{\omega }^{2}R$
• D. $\frac{GmM}{R^2}+m{\omega }^{2}R$

Answer 1

The correct option is C

$\frac{GmM}{R^2}-m{\omega }^{2}R$

The acceleration due to gravity at a place on the surface of the earth is given by
$g^{\prime }=g-{\omega }^{2}Rcos\varphi$ where $\varphi$ is the latitude of the place.
At the equator, $\varphi =0$. Therefore, $g^{\prime }=g-{\omega }^{2}R$
Therefore, the tension in the string will be $T=m{g}^{\prime }=mg-m{\omega }^{2}R=\frac{GmM}{R^2}-m{\omega }^{2}R$
Hence, the correct choice is (c).