Turn Safely

This problem can be solved by either using the ground frame or the bus frame.

Suppose we're looking at the ground frame ,

When the bus goes in a circular path, the friction acts towards the center and provides the necessary centripetal acceleration. Thus, the friction is $f=\dfrac{mv^2}{R}=m \omega^2 R$ . This friction has a tendency of toppling the bus in the counter-clockwise direction as shown in the diagram above.

If the speed of the bus is increased keeping the radius of the turn constant, the requirement of the centripetal force and hence the friction will increase. This will increase the chance of tipping.

If the time it takes to make the turn remains constant then the angular velocity also remains constant. Now, if the turning radius is increased then again the requirement of the centripetal force and hence the friction will increase. This will increase the chance of tipping.

If all the passengers sit on the upper deck, the center of mass of the bus will be higher and the same friction will create a greater turning moment. This will increase the chance of tipping.

Lastly, if the turning radius is increased and the speed remains constant then by the formula $f=\dfrac{mv^2}{R}$ , the friction decreases and hence the chance of toppling is decreased .

Now, suppose we are looking at the bus frame ,

In the bus frame, a pseudo force acts radially outwards on the center of mass of the bus. This is known as the centrifugal force. Now, the torque of the friction and the normal reaction are both zeroes about an axis passing through $A$ and going along the length of the bus. The torque of the centrifugal force and the weight of the bus are in opposite directions. The torque of the centrifugal force tries to topple the bus whereas the torque the weight of the bus stabilizes it. The factors that will increase the torque due to the centrifugal force will increase its chances toppling. Thus, if the speed remains constant, but the turning radius was increased, then the chance of tipping will be decreased.

When moving in a circle, the resultant centripetal force is $F=\frac{mv^2}{r}$ .

If we increase the turning radius, $r$ , but keep the speed constant, we have decreased the centripetal force, giving the bus less chance of toppling.

It is worth noting that moving passengers to the top deck would heighten the centre of mass, making the bus less stable.

Increasing the speed with constant radius would increase the centripetal force, making the bus more likely to tip.

Keeping the time constant but increasing the radius would give the bus more distance to cover in the same amount of time. Thus, the speed and the centripetal force would increase.

If the bus moves fast then as the centre of gravity (COG) of a double decker bus is situated quite above the ground so on slight sifting of COG from the middle vertical axis it generates enough torque to turn over the bus.This is also because of pseudo which helps in shifting of COG as it always acts away from the direction of motion.If passengers are more at top the COG is shifted further up which provides greater stability.But in 2nd case the bus moves at constant speed so there is no pseudo force as acceleration is 0