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Step-by-Step Solution
Step 1: Understand the problem
We are given:
• The earth’s radius, $R = 6400 \times 10^3 \,\text{m}$.
• Acceleration due to gravity, $g = 10 \,\text{m/s}^2$.
• When the earth is not rotating, the weight of a person at the equator is $W$.
• We want the earth to rotate so that this person’s weight becomes $\tfrac{3}{4}W$.
Step 2: Express the effective weight on a rotating Earth
When the person stands at the equator of a rotating Earth, the centripetal force required for circular motion is provided by a portion of the gravitational force. Hence, the effective weight $W^\prime$ is given by
$W^\prime = mg^\prime = mg - m\omega^2 R$,
where $\omega$ is the angular speed of Earth’s rotation, and $R$ is the Earth’s radius (cosine term simplifies to 1 because $\theta = 0^\circ$ at the equator).
Step 3: Relate the reduced weight to the angular speed
According to the problem, the new weight $W^\prime$ should be $\tfrac{3}{4}W$, i.e.,
$\tfrac{3}{4}mg = mg - m\omega^2 R.$
Simplifying,
$mg - \tfrac{3}{4}mg = m\omega^2 R \quad \Rightarrow \quad \tfrac{1}{4}mg = m\omega^2 R.$
Canceling $m$ from both sides, we get
$\tfrac{1}{4}g = \omega^2 R.$
Hence,
$\omega^2 = \dfrac{g}{4R} \quad \Longrightarrow \quad \omega = \sqrt{\dfrac{g}{4R}}.$
Step 4: Substitute numerical values
Given $g = 10\,\text{m/s}^2$ and $R = 6400 \times 10^3\,\text{m}$, substitute these into the expression for $\omega$:
$\omega = \sqrt{\dfrac{10}{4 \times 6400 \times 10^3}}.$
First simplify the denominator:
$4 \times 6400 \times 10^3 = 25600 \times 10^3 = 2.56 \times 10^7.$
Thus,
$\omega = \sqrt{\dfrac{10}{2.56 \times 10^7}}.$
Now compute the numerical value:
$\omega \approx \sqrt{3.90625 \times 10^{-7}} \approx 0.63 \times 10^{-3}\,\text{rad/s}.$
Step 5: State the final answer
Therefore, the angular speed with which the Earth must rotate so that the person at the equator weighs $\tfrac{3}{4}W$ is
$\boxed{0.63 \times 10^{-3}\,\text{rad/s}}.$
