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A road is 10 metre wide. Its radius of curvature is 50 metre. The outer edge is above the inner edge by a distance of 1.5 metre. This road is most suited for the velocity $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right.$ ]
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The correct answer is:
$8.5 \mathrm{~m} / \mathrm{s}$
Consider the free body diagram below:
By taking vertical and horizontal force balance:
$\mathrm{N} \sin \alpha=\mathrm{mv}^2 / \mathrm{R} \& \mathrm{~N} \cos \alpha=\mathrm{mg}$
The banking angle of the road can be written as: $\tan \alpha=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{\mathrm{h}}{\mathrm{w}}$
$\therefore \mathrm{v}=\sqrt{\frac{\mathrm{Rgh}}{\mathrm{w}}}=\sqrt{\frac{50 \mathrm{~m} \times 1.5 \mathrm{~m} \times 9.8 \mathrm{~m} / \mathrm{s}^2}{10 \mathrm{~m}}}=8.57 \mathrm{~m} / \mathrm{s}$
By taking vertical and horizontal force balance:
$\mathrm{N} \sin \alpha=\mathrm{mv}^2 / \mathrm{R} \& \mathrm{~N} \cos \alpha=\mathrm{mg}$
The banking angle of the road can be written as: $\tan \alpha=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{\mathrm{h}}{\mathrm{w}}$
$\therefore \mathrm{v}=\sqrt{\frac{\mathrm{Rgh}}{\mathrm{w}}}=\sqrt{\frac{50 \mathrm{~m} \times 1.5 \mathrm{~m} \times 9.8 \mathrm{~m} / \mathrm{s}^2}{10 \mathrm{~m}}}=8.57 \mathrm{~m} / \mathrm{s}$
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