What constant force must be applied tangentia | vedclass

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(A) The initial angular velocity of the Earth is $\omega_1 = \frac{2\pi}{T}$ rad/s,where $T = 86400 \, s$. The final angular velocity is $\omega_2 = 0

Vedclass · Accepted Answer

$ 1.3 	imes 10^{22} \, N $

Vedclass · Answer

(A) The initial angular velocity of the Earth is $\omega_1 = \frac{2\pi}{T}$ rad/s,where $T = 86400 \, s$. The final angular velocity is $\omega_2 = 0$. The time taken is $t = 86400 \, s$. The angular acceleration is $\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{0 - \frac{2\pi}{86400}}{86400} = -\frac{2\pi}{(86400)^2} \, rad/s^2$. The torque required is $	au = I\alpha$,where $I = \frac{2}{5}MR^2$ is the moment of inertia of the Earth. The tangential force $F$ is related to torque by $	au = FR$,so $F = \frac{I\alpha}{R} = \frac{2}{5}MR\alpha$. Substituting the values: $M = 6 	imes 10^{24} \, kg$,$R = 6.4 	imes 10^6 \, m$,and $\alpha = \frac{2\pi}{(86400)^2}$. $F = \frac{2}{5} 	imes (6 	imes 10^{24}) 	imes (6.4 	imes 10^6) 	imes \frac{2\pi}{(86400)^2} \approx 1.3 	imes 10^{22} \, N$.

What constant force must be applied tangentially at the equator to stop the Earth's rotation in one day?

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