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 $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right.$ ]

Question

Aicharya · Accepted Answer

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: $	an \alpha=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{\mathrm{h}}{\mathrm{w}}$ $	herefore \mathrm{v}=\sqrt{\frac{\mathrm{Rgh}}{\mathrm{w}}}=\sqrt{\frac{50 \mathrm{~m} 	imes 1.5 \mathrm{~m} 	imes 9.8 \mathrm{~m} / \mathrm{s}^2}{10 \mathrm{~m}}}=8.57 \mathrm{~m} / \mathrm{s}$

A road is 10 metre wide. Its radius of curvature is 50 metre. The outer edge is above the inner edge by a…

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