Question

A person stands on a spring balance at the equator.
(a) By what fraction is the balance reading less than his true weight?
(b) If the speed of earth’s rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

Answer 1

True weight is defined as the gravitational pull acting on anybody. A spring balance gives us the reading of the normal force acting on that body. On any object standing on earth a centrifugal force is applied because of the rotation of earth.

Complete answer:
True weight is defined as the gravitational pull acting on anybody. Whereas a spring balance gives us the reading of the normal force acting on that body. So,
True Weight $= mg$, in all the cases.
Where, $m =$mass of the person
$g =$acceleration due to gravity
But in this case, to get the readings of spring balance we need to know the normal force acting on the person, which is,
$\Rightarrow N + F = mg$
Where $N =$normal force
$\Rightarrow F =$Centrifugal force $= m{\omega ^2}R$
$\omega =$angular velocity of earth
$R =$Radius of the earth
So, $N + m{\omega ^2}R = mg$
$\Rightarrow N = mg - m{\omega ^2}R$
Now to calculate fractional change,
Fractional Change$= \dfrac{{TW - N}}{{TW}}$
Where $TW =$True Weight
So, Fractional change $= \dfrac{{mg - \left( {mg - m{\omega ^2}R} \right)}}{{mg}}$
$\Rightarrow \dfrac{{mg - mg + m{\omega ^2}R}}{{mg}}$
$\Rightarrow \dfrac{{m{\omega ^2}R}}{{mg}}$
Fractional change$= \dfrac{{{\omega ^2}R}}{g}$.................(1)
Since $\omega = \dfrac{{2\pi }}{T}$, where $T =$Time period of earth $= 24 \times 60 \times 60s$
$\Rightarrow R = 6400 \times {10^3}m$
$\Rightarrow g = 9.8m/{s^2}$
Substituting these values in equation one,
Fractional change $= \dfrac{{{{\left( {\dfrac{{2\pi }}{{24 \times 60 \times 60}}} \right)}^2} \times 6400 \times {{10}^3}}}{{9.8}}$
Fractional change $= 3.5 \times {10^{ - 3}}$
For second part of question, the angular velocity is changed to $\omega '$
$\Rightarrow N = mg - m{\omega '^2}R$
Since, $N = \dfrac{{mg}}{2}$ in this case, therefore,
$\Rightarrow \dfrac{{mg}}{2} = mg - m{\omega '^2}R$
$\Rightarrow m{\omega '^2}R = \dfrac{{mg}}{2}$
$\Rightarrow {\omega '^2}R = \dfrac{g}{2}$
Since $\omega = \dfrac{{2\pi }}{T}$, therefore,
$\Rightarrow \dfrac{{{{\left( {2\pi } \right)}^2}}}{{{T^2}}}R = \dfrac{g}{2}$
$\Rightarrow T = 2\pi \sqrt {\dfrac{{2R}}{g}}$
Substituting their values,
$\Rightarrow T = 2\pi \sqrt {\dfrac{{2 \times 6400 \times {{10}^3}}}{{9.8}}}$
$\Rightarrow T = 7175.768s$
$\Rightarrow T = 1.993hrs$

Note: The direction of the centrifugal force is always away from the centre of the axis of rotation. Whereas the direction of the centripetal force is always towards the centre with magnitude equal to that of centrifugal force.