Equilibrium of bicycle going in a circle

[serverxeon]

I have a few statements/questions, as stated below.
Would appreciate your comments on them.

1) The system is not in translational equilibrium because of a net leftwards force.
2) The system should be in rotational equilibrium based on logic;
But based on the FBD, if I take moments about the point of contact of the wheel with the floor, then only Weight in considered and the system should topple. But in reality we know it does not happen.

What have I overlooked?

Thanks!

 

[truesearch]

The cyclist is traveling in a circle...can you see which force is providing the centripetal force?

 

[serverxeon]

yea i know the friction force is providing the centripetal force.

but in a separate question,
if I consider the bicycle to be in rotational equilbrium,
why does the net moment abt any point not equate?

 

[rcgldr]

if I take moments about the point of contact of the wheel with the floor, then only Weight in considered and the system should topple.

Try using a different point to calculate the moments or torques involved.

 

[serverxeon]

My question is, why can I not take pivot about any point? Since the system is in rotational equilibrium.

If the answer only holds for one specific pivot point, then the system must not be in rotational equilibrium!

 

[rcgldr]

My question is, why can I not take pivot about any point? Since the system is in rotational equilibrium.

If using the contact point as the pivot point, the net torque is affected by inertial force, so you need to include the outward inertial force at the center of mass, and this is what I have seen in diagrams of a bicycle taking a turn. Using the contact point as the pivot point, then N and F don't generate torques, and the net torque due to W and the outwards inertial force (= -F) at the center of mass would be zero.

It would seem the only way to eliminate the torque due to inertial force is to use the center of mass as the pivot point. If you use the center of mass as the pivot point then W and the outwards intertial force don't generate torques, and the net torque due to N and F would be zero.

 

[PeterO]

I have a few statements/questions, as stated below.
Would appreciate your comments on them.

1) The system is not in translational equilibrium because of a net leftwards force.
2) The system should be in rotational equilibrium based on logic;
But based on the FBD, if I take moments about the point of contact of the wheel with the floor, then only Weight in considered and the system should topple. But in reality we know it does not happen.

What have I overlooked?

Thanks!

When we arrive at a contradiction, it usually means at least one of the original assumptions is incorrect.

The only assumption seems to be "The system should be in rotational equilibrium based on logic" so I expect that to be wrong.

What is the error?

Perhaps the translational and rotational equilibrium you refer to are for stationary objects only, or at least objects in an intertial (non-accelerating) frame of reference.

A bicycle being ridden in circle is in an accelerated frame of reference, so those equilibrium ides presumably don't apply.

 

[PeterO]

I have a few statements/questions, as stated below.
Would appreciate your comments on them.

1) The system is not in translational equilibrium because of a net leftwards force.
2) The system should be in rotational equilibrium based on logic;
But based on the FBD, if I take moments about the point of contact of the wheel with the floor, then only Weight in considered and the system should topple. But in reality we know it does not happen.

What have I overlooked?

Thanks!

I think the reason is as simple as:

The force acting through the tyre has to have an upward component to balance weight, and an inward (left in the picture) component to provide the required centripetal force.
The net Force on the tyre is thus up and to the left.
If the rider is to be stable, that force must act through the centre of mass, so the cyclist has to lean his bike over so that the centre of mass is on that line of action.

Note: if you have ever ridden a bike, you would realize that you can just hang your body off the side of the bike so that, although the bike frame is vertical, the centre of mass of you and the bike is in a appropriate location [up and left]. That can usually only be done when turning slowly, or tracing out a large radius circular path.