Why does reaction force change when a cyclist 'bends'?

Question

I am learning about the 'bending' of a cyclist

I was told that the reaction force also 'tilts' when the cyclist is tilting (like in the figure), How is this possible? Isn't the reaction force always normal to the cyclist?

Answer 1

well it is just that reaction force is not always normal to the plane.Instead the normal force is always normal to the plane. The reaction force is composed of two parts

1. The normal force which is normal to the surface i.e, the $ R\cos(\theta)$
2. And the friction force which is horizontal i.e, the $ R\sin(\theta)$

Answer 2

I think $R$ is meant to be the force exerted by the cycle on the cyclist. The confusing thing about the diagram is that it draws the cyclist and cycle as one object, but then shows $R$, which is an internal force as far as this combined object is concerned. The diagram also omits the equal and opposite force $-R$ which the cyclist exerts on the cycle, the normal force that the road exerts on the cycle, and the cycle's weight.

It would be much clearer to draw two separate free-body diagrams - one for the cyclist, with forces $R$ from the cycle and the cyclist's weight $mg$; and one for the cycle with forces $-R$ from the cyclist, the cycle's weight $Mg$, friction from the road surface (horizontal) and the normal force from the road surface (vertical).

Answer 3

I was told that the reaction force also 'tilts' when the cyclist is tilting (like in the figure), How is this possible?

You are right to question this statement, as it is not always true.

First lets establish what the reaction force is in this context. It is the resultant of the normal force of the road surface and the friction force providing the centripetal force (red dashed vectors in the diagram below). When a cyclist is performing a turn in a stable manner this reaction force is aligned with the plane of symmetry of the frame. In the diagram you posted, the bike is not stabilised and there is anti-clockwise torque present because the gravitational force acting at the com is not balanced by the vertical component of the reaction force ($R \ \cos \theta$) and because the friction force at the road surface is greater than the centripetal force ($R sin \theta = mR\omega^2$) acting at the com. In the following diagram I have adjusted the vectors for the situation where the bike is in a stable turn.

The green vectors are components of the reaction force and vertical component balances the weight and the horizontal component is unbalanced and provides the centripetal force to accelerate the com inwards as is required for circular motion.

An additional free body diagram has been done at the contact point with the road which is basically the same as the free body diagram analysed at the com. It can be seen that the friction force at the road surface now matches centripetal force at the com and there is no torque acting on the system.

If the cyclist leans over further and adjust the front wheel angle so that the radius of the turn remains the same, then the reaction force continues to point in the same direction but is now not aligned with the bike frame. This results in an imbalance causing a clockwise torques and the bike to topple to the ground if the situation is not corrected by the rider. It is the actions of a competent rider that aligns the bike lean angle with the reaction force to keep things in balance. The design of the rake angles, fork angles and frame geometry and natural gyroscopic forces of the wheels tend to keep a bike stabilised in a steady turn, without any active input from the rider, but if the rider is a learner and leans the bike to the wrong angle, the reaction force (as defined earlier) will not automatically compensate and they will probably fall off.

If a competent cyclist encounters a tightening turn or speeds up or slows down, they adjust the lean angle of the bike and the turn angle of the front wheel in ensure that the reaction force stays inline with the bike frame.

All the above makes the simplifying assumption that cyclist keeps his com in line with the bike frame at all times.

Just to be clear, if the cyclist in a stable turn leans the bike to a different angle and does not adjust the handlebar accordingly, the turn radius will change and with it the reaction force will also change, but not necessarily keeping in line with the frame.

Answer 4

Most probably I didnt get your question. What do you mean by Reaction force is it Normal force. If yes then the normal always acts perpendicular it is just that we have spitted it into components of force (R cos-theta, R sin-theta) which can be compared to other components of force in the same direction(or same axis)