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4-1.Newton's Laws of Motion
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A vehicle is moving with a velocity $v$ on a curved road of width $b$ and radius of curvature $R.$ For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is
A$v^2b / Rg$
B$vb / Rg$
C$vb^2 / Rg$
D$vb / R^2g$
Solution
We know that for the vehicle banking
on a curved path, $\tan \theta=\frac{v^{2}}{r g}$
Here, the vehicle moves with velocity
$v$ and radius of curvature $R$ and width is b.
So, here $\theta$ is small (refer figure) and thus, $\sin \theta=\frac{h}{b}=\tan \theta$
Now, we can equate $\frac{h}{b}=\frac{v^{2}}{R g}$
$\Rightarrow h=\frac{b v^{2}}{R g},$ will be the required elevation between the outer and inner edges of the road.
on a curved path, $\tan \theta=\frac{v^{2}}{r g}$
Here, the vehicle moves with velocity
$v$ and radius of curvature $R$ and width is b.
So, here $\theta$ is small (refer figure) and thus, $\sin \theta=\frac{h}{b}=\tan \theta$
Now, we can equate $\frac{h}{b}=\frac{v^{2}}{R g}$
$\Rightarrow h=\frac{b v^{2}}{R g},$ will be the required elevation between the outer and inner edges of the road.
Standard 11
Physics
