Question

If earth suddenly stops rotating, then the weight of an object of mass $m$ at equator will $(\omega$ is angular speed of earth and $R$ is its radius $)$

• A. decrease by $m{\omega }^{2}R$
• B. increase by $m{\omega }^{2}R$
• C. decrease by $m\omega R$
• D. increase by $m\omega R$

Answer 1

The correct option is B increase by $m{\omega }^{2}R$

The apparent weight of an object at any place on the Earth due to the effect of rotation of the Earth, is given by$$m{g}^{\prime }=m(g-{\omega }^{2}R{\cos }^2\lambda )$$ Where, $\lambda$ is the latitude,
$R$ is the radius and
$\omega$ is the rotational velocity.

At the equator, $\lambda =0^{\circ }$. So,

$m{g}_{equator}=(mg-m{\omega }^{2}R{\cos }^2{0}^{\circ })$

$\Rightarrow m{g}_{equator}=(mg-m{\omega }^{2}R)...\left(1\right)$

When, Earth suddenly stop rotating i.e., $\omega =0$,

$\Rightarrow m{g}_{equator}=(mg-m\times 0^2\times R)$

$\therefore m{g}_{equator}=mg...\left(2\right)$

Thus, from equation $\left(1\right)$ and $\left(2\right)$, It can be concluded that weight of an object of mass $m$ at equator will increase by $m{\omega }^{2}R$.

Hence, option (b) is the correct answer.

Why this question? To familiarize students with the change in acceleration due to gravity due to the rotation of the earth.