Question 359: Gravitation

Question

Consider a particle of mass $m$ suspended by a string at the equator. Let $R$ and $M$ denote radius and mass of the earth. If $\omega$ is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be $\left(\cos 0^{\circ}=1\right)$

Accepted Answer

When a body suspended by the string situated at position $P$ as shown in the figure, where latitude is $\lambda$, then body is also rotated with angular frequency $(\omega)$ of earth, hence tension on the string is given by
$$\begin{aligned}
& T=m g-m r \omega^2 \cos \lambda \
T & =m \cdot \frac{G M}{R^2}-m r \omega^2 \cos \lambda\left[	herefore g=\frac{G M}{R^2}ight] \
T= & \frac{G M m}{R^2}-m r \omega^2 \cos \lambda\quad	ext{.... (i)}
\end{aligned}$$
When body is suspended at equator, then
$$\lambda=0 	ext { and } r=R$$
$	herefore$ From Eq. (i), we have,
$$\begin{aligned}
& T=\frac{G M m}{R^2}-m R \omega^2 \cos 0^{\circ} \
& T=\frac{G M m}{R^2}-m R \omega^2
\end{aligned}$$

Answer

$\frac{G M m}{R^2}$

Answer

$\frac{G M m}{2 R^2}$

Answer

$\frac{G M m}{2 R^2}+m \omega^2 R$

Answer

$\frac{G M m}{R^2}-m \omega^2 R$

Gravitation

Consider a particle of mass m suspended by a string at the equator. Let R and M denote radius and mass of the earth. If \omega is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be \left(\cos 0^{\circ}=1\right)

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