JEE Main & Advanced Physics Two Dimensional Motion Overturning of Vehicle

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 \[v=\sqrt{\frac{gra}{h}}\]