Question
If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?
✓
Answer
in the diagram
$\text{R}\cos\theta=\text{mg}\ \dots(1)$
$\text{R}\sin\theta=\frac{\text{mv}^2}{\text{r}}\ \dots(2)$
Dividing equation (1) with equation (2),$\tan\theta=\frac{\text{mv}^2}{\text{rmg}}=\frac{\text{v}^2}{\text{rg}}$
$\text{v}=36\text{km}/\text{hr}=10\text{m}/\text{sec},\text{r}=30\text{m}$
$\tan\theta=\frac{\text{v}^2}{\text{rg}}$
$=\frac{100}{30\times10}=\frac{1}{3}$
$\Rightarrow\theta=\tan^{-1}\Big(\frac{1}{3}\Big)$
$\text{R}\cos\theta=\text{mg}\ \dots(1)$
$\text{R}\sin\theta=\frac{\text{mv}^2}{\text{r}}\ \dots(2)$
Dividing equation (1) with equation (2),$\tan\theta=\frac{\text{mv}^2}{\text{rmg}}=\frac{\text{v}^2}{\text{rg}}$
$\text{v}=36\text{km}/\text{hr}=10\text{m}/\text{sec},\text{r}=30\text{m}$
$\tan\theta=\frac{\text{v}^2}{\text{rg}}$
$=\frac{100}{30\times10}=\frac{1}{3}$
$\Rightarrow\theta=\tan^{-1}\Big(\frac{1}{3}\Big)$
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