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Question

Zorch, an archenemy of Rotation Man, decides to slow Earth's rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Rotation Man is not immediately concerned, because he knows Zorch can only exert a force of

4.00×107N4.00 × 10^7 N

(a little greater than a Saturn V rocket's thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Rotation Man time to devote to other villains.)

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First, we calculate the torque (τ)\left(\tau\right) produced by the force (F)\left(F\right). We also need to calculate the moment of inertia of Earth about its axis (IEarth)\left(I_\text{Earth}\right).

τ=REarthF=(6.37106)(4.00107)=2.551014 NmIEarth=25MEarthREarth2=25(5.951024)(6.37106)2=9.661037 kgm2\begin{align*} \tau&=- R_\text{Earth} F \\ &=- \left(6.37 \cdot 10^6\right) \left(4.00 \cdot 10^7\right) \\ &=-2.55 \cdot 10^{14} \text{ N}\cdot \text{m} \\ I_\text{Earth}&=\dfrac{2}{5} M_\text{Earth} {R_\text{Earth}}^2 \\ &=\dfrac{2}{5} \left(5.95 \cdot 10^{24}\right) \left(6.37 \cdot 10^6\right)^2 \\ &=9.66 \cdot 10^{37} \text{ kg}\cdot \text{m}^2 \end{align*}

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