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# Because of the friction between the water in oceans with the earth's surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by 0⋅0016 rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is 6⋅0 × 1024 kg.

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Solution

## Rate of change of angular velocity, i.e., angular acceleration,α = $\left(\frac{0.0016}{100}\right)\mathrm{rad}/\mathrm{day}$$⇒\alpha =\left\{\frac{0.0016}{{\left(86400\right)}^{2}×100×365}\right\}\left[1\mathrm{year}=365\mathrm{days}=365×86400\mathrm{sec}\right]$Torque produced by the ocean water in decreasing the Earth's angular velocity,$\tau =I\alpha =\frac{2}{5}m{r}^{2}\alpha \phantom{\rule{0ex}{0ex}}=\frac{2}{5}×6×{10}^{24}×{\left(64×{10}^{5}\right)}^{2}×\left\{\frac{0.0016}{{86400}^{2}×100×365}\right\}\phantom{\rule{0ex}{0ex}}=5.8×{10}^{20}\mathrm{N}-\mathrm{m}$  Suggest Corrections  0      Similar questions
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