Overturning of Vehicle
Category : JEE Main & Advanced
When a car moves in a circular path with speed more than a certain maximum speed then it overturns even if friction is sufficient to avoid skidding and its inner wheel leaves the ground first
Weight of the car = mg
Speed of the car = \[\upsilon \]
Radius of the circular path = r
Distance between the centre of wheels of the car = 2a
Height of the centre of gravity (G) of the car from the road level = h
Reaction on the inner wheel of the car by the ground \[={{R}_{1}}\]
Reaction on the outer wheel of the car by the ground \[={{R}_{2}}\]
When a car move in a circular path, horizontal friction force F provides the required centripetal force
i.e., \[F=\frac{m{{v}^{2}}}{R}\] ...(i)
For rotational equilibrium, by taking the moment of forces \[{{R}_{1}},\,\,{{R}_{2}}\] and F about G
\[Fh+{{R}_{1}}a={{R}_{2}}a\] ...(ii)
As there is no vertical motion so \[{{R}_{1}}+\,{{R}_{2}}=mg\] ...(iii)
By solving (i), (ii) and (iii)
\[{{R}_{1}}=\frac{1}{2}M\left[ g-\frac{{{v}^{2}}h}{ra} \right]\] ...(iv)
and \[{{R}_{2}}=\frac{1}{2}M\left[ g+\frac{{{v}^{2}}h}{ra} \right]\] ...(v)
It is clear from equation (iv) that if \[\upsilon \] increases value of \[{{R}_{1}}\] decreases and for \[{{R}_{1}}=0\]
\[\frac{{{v}^{2}}h}{ra}=g\] or \[v=\sqrt{\frac{gra}{h}}\]
i.e. the maximum speed of a car without overturning on a flat road is given by \[v=\sqrt{\frac{gra}{h}}\]
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