Question

If the Earth has no rotational motion, the weight of a person on the equator is $W$ . Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $\dfrac{3}{4}W$ . Radius of the Earth is $6400km$ and $g = 10m{s^{ - 2}}$.
(A) $1.1 \times {10^{ - 3}}rad/s$
(B) $0.63 \times {10^{ - 3}}rad/s$
(C) $0.28 \times {10^{ - 3}}rad/s$
(D) $0.83 \times {10^{ - 3}}rad/s$

Answer 1

Hint: We are given with the weight of the person on the equator when the earth has no rotational motion and we asked to find the angular velocity of the earth such that due to this, the weight of the person on the equator becomes ${\dfrac{3}{4}^{th}}$ of the original value. Thus, we will find the new acceleration due to gravity due to the rotation. Then finally, we will calculate the angular velocity out of that.
Formulae used
$g' = g - {\omega ^2}R{\cos ^2}\theta $
Where, $g'$ is the new acceleration due to gravity, $g$ is the original acceleration due to gravity, $\omega $ is the angular velocity, $R$ is the radius of the planet and $\theta $ is the angle between the equator of the planet and the planet’s axis of rotation.

Step By Step Solution
Here,
The original weight of the person is $W$.
Also, weight of a body on a planet is given by $mg$, where $m$ is the mass of the body.
Thus,
$W = mg$
Now,
For the weight after the rotational motion of earth,
$W' = mg'$
But, according to the question,
$W' = \dfrac{3}{4}W$
Thus, we get
$mg' = \dfrac{3}{4}mg$
After cancellation of $m$, we get
$g' = \dfrac{3}{4}g$
Also,
$g' = g - {\omega ^2}R{\cos ^2}\theta $
Here,
$\theta \approx {0^c}$ as we the axis of rotation of earth and the equator of the earth are very close to each other.
Thus, we get the formula to be
$g' = g - {\omega ^2}R$
Now,
Putting in $g' = \dfrac{3}{4}g$ , we can say
$\dfrac{g}{4} = {\omega ^2}R$
Now,
Substituting the value $R = 6.4 \times {10^6}m$ and $g = 10m{s^{ - 2}}$, we get
${\omega ^2} = \dfrac{{10}}{{25.6 \times {{10}^6}}} = \dfrac{1}{{25.6}} \times {10^{ - 5}} = \dfrac{1}{{256}} \times {10^{ - 4}}$
Thus,
Taking the square on the angular velocity to the other side of the equation, we get
$\omega = \dfrac{1}{{16}} \times {10^{ - 4}} = 0.063 \times {10^{ - 4}} = 0.63 \times {10^{ - 3}}rad/s$

Hence, the answer is (B).

Note: We have taken $\theta \approx {0^c}$ as in the case of earth, the axis of rotation and the equator lie very close to each other. But, if in case of any other planet, $\theta $ may or may not be zero. But the formula for the new acceleration due to gravity remains the same only the calculations differ.