Question 251: Work, Energy and Power & Circular Motion

Question

A railway track is banked for a speed ',$$v$$' by elevating outer rail by a height '$$h$$' above the inner rail. The distance between two rails is 'd' then the radius of curvature of track is ( $$\mathrm{g}=$$ gravitational acceleration)

Accepted Answer

From figure,
$$\begin{aligned}
& 	an 	heta=\frac{\mathrm{h}}{\mathrm{d}} \
	herefore \quad & \frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{\mathrm{h}}{\mathrm{d}} \quad \ldots .\left(\because 	an 	heta=\frac{\mathrm{v}^2}{\mathrm{rg}}ight) \
	herefore \quad & \mathrm{r}=\frac{\mathrm{v}^2 \mathrm{~d}}{\mathrm{gh}}
\end{aligned}$$

Answer

$$\frac{\mathrm{v}^2 \mathrm{~d}}{\mathrm{gh}}$$

Answer

$$\mathrm{\frac{2 v^2}{g d h}}$$

Answer

$$\mathrm{\frac{g d}{2 v^2 h}}$$

Answer

$$\mathrm{\frac{v^2}{2 g h d}}$$

Work, Energy and Power & Circular Motion

A railway track is banked for a speed ',$v' by elevating outer rail by a height 'h' above the inner rail. The distance between two rails is 'd' then the radius of curvature of track is ( \mathrm{g}=$ gravitational acceleration)

Choose the correct answer: