Question

The speed with which the earth has to rotate on its axis so that a person on the equator would weigh $\dfrac{3}{5}th$ as much as present will be. [Take a equilateral radius of earth =6400 km]
a. $3.28 \times {10^{ - 4}}rad/\sec$
b. $7.28 \times {10^{ - 3}}rad/\sec$
c. $3.28 \times {10^{ - 3}}rad/\sec$
d. $7.826 \times {10^{ - 4}}rad/\sec$

Answer 1

Apparent weight of a person on the equator is the difference between the actual weight and the centrifugal force created for the rotation of the earth.

Formula used:
Write down the Centrifugal force for earth’s rotation:
${F_C} = m{\omega ^2}R$ -------(1)
Where,
${F_C}$is the centrifugal force for earth’s rotation,
m is the mass of the person,
$\omega$is the angular velocity of earth,
R is the radius of earth at the equator.

Weight of a person:
$W = mg$ ------(2)
Where,
W is the weight of the person,
g is the acceleration due to gravity of earth.

Apparent weight at the equator:
$W' = W - {F_C}$ ------(3)
Where,
$W'$is the apparent weight at the equator.

Complete step by step answer:
Given:
Radius of earth, $R = 6400km = 6.4 \times {10^6}m$,
Weight of the person will be $\dfrac{3}{5}th$ of the present i.e.:
$W' = \dfrac{{3W}}{5}$ ------(4)
Also, assume $g = 9.8m{s^{ - 2}}$
To find: Angular velocity of earth i.e. $\omega$.

Step 1:
Substitute eq(1), eq(2) and eq(4) in eq(3) to get the expression for $\omega$as:
$\dfrac{{3W}}{5} = W - m{\omega ^2}R \\\ \Rightarrow m{\omega ^2}R = \left( {1 - \dfrac{3}{5}} \right)W \\\ \Rightarrow m{\omega ^2}R = \dfrac{2}{5}mg \\\ \Rightarrow {\omega ^2} = \dfrac{{2g}}{{5R}} \\\ \Rightarrow \omega = \sqrt {\dfrac{{2g}}{{5R}}} \\\$ -------(5)

Step 2:
Put the given values of g and R in the eq(5) to get:

\omega = \sqrt {\left( {\dfrac{2}{5} \times \dfrac{{9.8}}{{6.4 \times {{10}^6}}}} \right)} rad/\sec \\\ = 7.826 \times {10^{ - 4}}rad/\sec \\\ \end{gathered} $$ **Hence, the correct answer is option (D).** **Note:** Centrifugal force arises for any type of rotational motion. In eq(1) we see that this force is dependent on the radius of Earth. But this is true only for any point on the equator. In general, for any arbitrary point P on earth the term R in eq(1) is replaced by r where r is the radius of the horizontal circle containing the point P and centering at some point on the axis. You can express r in terms of R as: $r = R\cos \lambda $ where, $\lambda $ is the latitude of that point. Now, you can clearly notice that at equator i.e. for $\lambda = {0^ \circ }$, $r = R$ and at the poles i.e. for $\lambda = {90^ \circ }$, $r = 0$. So, the centrifugal force is maximum at the equator and 0 at the poles.