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(A) The weight of a man at the equator is given by $w_{eq} = m(g - \omega^2 R)$,where $m$ is the mass,$g$ is the acceleration due to gravity,$\omega$

Vedclass · Accepted Answer

Increase by $0.34$

Vedclass · Answer

(A) The weight of a man at the equator is given by $w_{eq} = m(g - \omega^2 R)$,where $m$ is the mass,$g$ is the acceleration due to gravity,$\omega$ is the angular velocity of the Earth,and $R$ is the radius of the Earth.At the poles,the effect of rotation is zero,so the weight is $w_p = mg$.The change in weight is $\Delta w = w_p - w_{eq} = mg - m(g - \omega^2 R) = m\omega^2 R$.The fractional change in weight is $\frac{\Delta w}{w_{eq}} = \frac{m\omega^2 R}{m(g - \omega^2 R)} = \frac{\omega^2 R}{g - \omega^2 R}$.Given $\omega^2 R \approx 0.0337 \, m/s^2$ and $g \approx 9.81 \, m/s^2$,we have:$\frac{\Delta w}{w_{eq}} = \frac{0.0337}{9.81 - 0.0337} \approx \frac{0.0337}{9.7763} \approx 0.003447$.Converting to percentage: $0.003447 	imes 100 = 0.3447 \%$.Thus,the weight increases by approximately $0.34 \%$.

During the motion of a man from the equator to the pole of the Earth,his weight will ....... $\%$ (neglect the effect of change in the radius of the Earth).

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