Phy-overturning of a car

With reference to the diagram below:

圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Sep10/Crazymech1.jpg


Suppose that friction = f for centripetal force.

Then, taking moment about the centre of gravity of the car, we have:

R1 provides an anti-clockwise moment aR1

R2 provides a clockwise moment aR2

f provides a clockwise moment fh.

Now, when overturning just begins, the left tyres just lost contact with the ground, i.e. R2 = 0

Hence considering rotational equilibrium:

aR1 = fh

f = aR1/h

But now, since R2 = 0, R1 should support the whole weight of the car and therefore:

f = amg/h

Moreover, f is centripetal force mv2/r and hence:

mv2/r = amg/h

v2 = rag/h

Note: m is the mass of the car

This is a very standard example, in fact, the solution can be found in most physics textbooks.

Let F1 and F2 be the respective frictional forces at the two wheels of the car.
Then, for the car moves in a circular track of radius r, we have, horizontally,
F1 + F2 = mv^2/r ----------------------- (1)
where m is the mass of the car and v is its speed
In the vertical direction, R1 + R2 = mg ------------- (2)
where g is the acceleration due to gravity

Also, taking moment about the centre of gravity,
(F1 + F2).h + (R2).a = (R1).a --------------- (3)

Substitute (1) and (2) into (3),
mv^2.h/r + (mg-R1)a = (R1).a
solve for R1 gives R1 = (m/2).[g + (v^2.h)/(ra)]
and R2 = (m/2).[g - (v^2.h)/(ra)]

We can see that R1 is always positive irrespective of the speed v.
but R2 will become zero when (v^2.h)/(ra) = g
i.e. v^2 = gra/h
This is the condition when the car just begins to overturn when the normal reaction R2 at the inner wheels becomes zero.