Question

The radius of the earth is $6400km$ and $g=10m/s^2$. In order that a body of $5kg$ weighs zero at the equator, the angular speed of the earth is:

• A. $1/80rad/s$
• B. $1/400rd/s$
• C. $1/800rad/s$
• D. $1/1600rad/s$

Answer 1

The correct option is C $1/800rad/s$

$\begin{array}{cc}\text{Given}-& \ast \text{Radius of earth}R=6400km\\ & \ast \text{acceleration due to growity}g=10m/s^2\\ & \ast \text{mass of the Body}=5kg\end{array}$

$\begin{array}{c}\text{To get an apparent weight -}\\ \text{The weight of the Body must be counter}\\ \text{balanud by the centrepetal forct due to}\\ \text{the Rotation of earth.}\\ \text{weight = centripetal force}\end{array}$

$\begin{array}{c}\text{We Rnow that centrepetal foru is}\\ \begin{array}{ccc}cF& =\frac{m{v}^2}{r}=m{w}^{2}r& (v=rw)\text{for circular motion}\\ \text{where}v=\text{velocity ,}\omega =\text{angular velocity}& & \end{array}\end{array}$

$\begin{array}{c}\Rightarrow mg=m{\omega }^2\mu \\ \Rightarrow \omega =\sqrt{\frac{g}{r}}=\sqrt{\frac{10}{6400\times 1000}}\\ \omega =\frac{1}{800}\text{read}18\\ \text{Hence, option (C) is correct.}\end{array}$