Question

Because of the Earth's rotation, a plumb bob does not hang exactly along a line directed to the center of the Earth. How much does the plumb bob deviate from a radial line at$35.0^0$north latitude? Assume the Earth is spherical.

Short answer

$\mathrm{\gamma }=0.0928°$

Step-by-step solution

Given data

The deviation of the bob is

Concept introduction

The centripetal force will act inward, causing centripetal acceleration.

The equation for the centripetal acceleration is as follows:

$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}}$

Here, v is the velocity, and r is the radius.

Draw figure and calculate acceleration

Draw the situation.

The forces are:

Centripetal Force ${\mathrm{F}}_{\mathrm{c}}$.

Weight W.

Tension T.

$\begin{array}{c}\mathrm{r}=6.37\times {10}^6\cos 35°\\ \approx 5.22\times {10}^6\mathrm{m}\end{array}$

$$\begin{gathered} \mathrm{Calculate}\mathrm{radius} \\ \mathrm{r}={\mathrm{R}}_{\mathrm{e}}\mathrm{cos\varphi } \\ {\mathrm{R}}_{\mathrm{e}}=6.37\times {10}^6\mathrm{m} \\ \mathrm{\varphi }=35° \end{gathered}$$

$$\begin{gathered} T=24\mathrm{hr}\left(\frac{86400s}{1hr}\right)=86400\mathrm{s} \\ \mathrm{v}=\frac{2\mathrm{\pi r}}{\mathrm{T}} \\ =\frac{2\mathrm{\pi }\times 5.22\times {10}^6\mathrm{m}}{86400\mathrm{s}} \\ \approx 379.46\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} a=\frac{v^2}{r} \\ \frac{{\displaystyle {\left({\displaystyle 379.46m/s}\right)}^2}}{{\displaystyle 5.22\times {10}^{6}m}} \\ \approx 0.0276\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

To Find Equation

$\mathrm{Tsin\gamma }={\mathrm{F}}_{\mathrm{c}}\mathrm{sin\varphi }$ ...(1)

Net force in x-direction and y-direction

$$\begin{gathered} \mathrm{Tcos\gamma }=\mathrm{W}-{\mathrm{F}}_{\mathrm{c}}\mathrm{cos\varphi } \\ \sum {\mathrm{F}}_{\mathrm{x}}=0 \end{gathered}$$...2

$$\begin{gathered} \tan \gamma =\frac{F_c\sin \varphi }{W-F_c\cos \varphi } \\ \tan \gamma =\frac{ma\sin {35}^0}{{\displaystyle mg-ma\cos {35}^0}} \\ \tan \gamma =\frac{a\sin \varphi }{g-a\cos \varphi } \end{gathered}$$

$$\begin{gathered} \tan \gamma =\frac{{\displaystyle (0.0276m/s^2)Sin{35}^0}}{9.81m/s^2-\left(0.0276m/s^2\right)Cos{35}^0} \\ \approx 1.61746\times {10}^{-3} \end{gathered}$$

$\gamma =0.09267$