Question

what constant force ,tangential to the equator should be applied to the earth to stop its rotation in one day ​

Answer 1

Answer:

The constant force ,tangential to the equator should be applied to the earth to stop its rotation in one day ​is $F = 1.29 \times 10^4 N$

Explanation:

As we know that the moment of inertia of the Earth is given as

$I = \frac{2}{5}MR^2$

now we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.4 \times 10^6 m$

now we have

$I = \frac{2}{5}(5.98 \times 10^6)(6.4 \times 10^6)^2$

$I = 9.8 \times 10^{19} kg m^2$

now angular speed of the Earth is given as

$\\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

now angular acceleration of the earth to stop it in one day

$\alpha = \frac{\omega_f - \omega_i}{T}$

$\alpha = \frac{7.27 \times 10^{-5}}{24 \times 3600}$

$\alpha = 8.42 \times 10^{-10} N$

now we have

$F(6.4 \times 10^6) = (9.8 \times 10^{19})(8.42 \times 10^{-10})$

$F = 1.29 \times 10^4 N$

#Learn

Topic : Torque

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