Why Taking Torques About the CG Matters for Motorcycle Riders

[lonewolf5999]

This is not really a homework problem per se, it's just a question I thought of when attempting a circular motion problem.

Imagine a motorcycle rider negotiating a bend on a level road. The centripetal force is then provided by lateral friction between the motorcycle's wheels and the road. At the same time, in order to avoid toppling into the bend, the rider must tilt his motorcycle into the bend. The condition for rotational equilibrium is net torque = 0, where moments are taken about the centre of gravity of the rider.

By taking moments as such, we find that the lateral friction acting on the motorcycle wheels will provide a torque sufficient to counteract the torque acting in the opposite direction due to the normal reaction of the road on the motorcycle.

But why take moments about the c.g. ?

If you take moments about the point of contact of the motorcycle with the road, for example, you'd find that only one force creates a torque about that axis - the weight of the motorcycle, acting vertically downwards through the c.g. of the rider and cycle. In this position wouldn't the condition for rotational equilibrium be violated?

So yeah my question is basically why torques must be taken about the c.g. , and why taking torques about any other point in this situation won't work.

I apologise for the wordiness. If my explanation of my problem is unclear I'll try to upload a picture of what I'm talking about. Thanks!

 

[Dick]

You don't HAVE to take moments about c.g. Other points will work just fine. If the moments cancel around one point then they will cancel around all other points as well. You just may have to work harder to realize this.

 

[lonewolf5999]

Ok try this. There are only three forces acting on the rider and his cycle: lateral friction pointing towards the centre of the circle, normal reaction force vertically upwards (both forces act at the point of contact between the cycle wheels and the ground), and the weight of the cycle acting vertically downwards through the rider's c.g.

Taking moments about the point of contact of the wheel with the ground, the frictional force and normal reaction force have no moment arm and so contribute zero torque. Only the weight of the rider contributes a torque, and since there is no other torque to counter it, the rider cannot be in rotational equilibrium.

Where has my analysis gone wrong? Am I omitting some other force?

 

[Dick]

That would be correct except that normal force cancels gravity so there is nothing to cancel the friction reaction force pushing the cycle forward. So the point you are calculating moments around is accelerating. That makes things difficult to interpret. Luckily there is another force. Air friction.

 

[kudoushinichi88]

I think it's because the center of gravity moves when you tilt you bike when you turn due to the weight distribution. Though, I'm not sure how to explain it mathematically yet.

 

[Dick]

Just to clarify, the notion that if torques cancel when computed around one point, then they cancel around any other point is dependent on the sum of the forces being zero. Imagine an object with only one force acting on it.

 

[kudoushinichi88]

Oh... if air force plays a role in this problem, does that mean that if we are riding in vacuum, we wouldn't need to tilt our bodies when we turn?

 

[Dick]

Oh... if air force plays a role in this problem, does that mean that if we are riding in vacuum, we wouldn't need to tilt our bodies when we turn?

Why not? I'm not sure you are really thinking about the forces involved.

 

[lonewolf5999]

So the point you are calculating moments around is accelerating. That makes things difficult to interpret. Luckily there is another force. Air friction.

Could you elaborate on what you mean by "difficult to interpret", and the implications this has on trying to take moments about the point of contact of the motorcycle wheels with the ground?

Also, considering that air friction acts to oppose the motion of the rider, it would be acting in the direction opposite to the tangential velocity of the rider. Thus there wouldn't be any way for air friction to be counteracting the torque generated by the weight of the rider about the point of the contact of the wheels with the ground. This is because the torque generated by air friction about the point of contact of the wheels with the ground acts in a different plane from that generated by the weight about the said point. Thus air friction does not help explain the supposed problem of the rider being in rotational disequilibrium. Is this right?

The point I am trying to make here is: how do I explain the problem of the rider being in rotational disequilibrium if we take moments about that contact point. I've been checking a few textbooks with worked examples, but none of them seem to offer any explanation as to how to resolve this problem.

 

[Dick]

Could you elaborate on what you mean by "difficult to interpret", and the implications this has on trying to take moments about the point of contact of the motorcycle wheels with the ground?

I mean that you can only choose an arbitrary point to calculate the torque if the sum of the forces is zero, otherwise you have to use the axis of rotation. I don't follow your argument in the second paragraph that friction and weight operate in 'different planes'. Regarding the wheel as the axis, weight pulls the rider forward and friction pushes him backwards. They can perfectly well cancel.

 

[lonewolf5999]

I mean that you can only choose an arbitrary point to calculate the torque if the sum of the forces is zero, otherwise you have to use the axis of rotation.

Aha! This is what I was actually asking when I asked why taking torques about any axis other than the c.g. doesn't yield a result which shows the rider in rotational equilibrium.

But how can we know that the rider will tend to rotate about his c.g.? Why can't he rotate about any other point, say again, the point of contact between his wheel and the ground?

About the friction thing - I didn't realize that you were talking about air friction acting to oppose the rider from toppling inwards. My mistake for misinterpreting what you were trying to say.

Thanks for your help so far.

 

[Dick]

The point is that if there is no rotation around one axis, then there is no rotation around another. That's why in equilibrium you can chose any axis and get the same answer. If the rider is rotating and accelerating then you have a dynamics problem. It's much more complicated. You can choose any coordinate system you want, but you can't apply the same simplifications that you can in the equilibrium case. I.e. he does rotate around other points as well.

 

[lonewolf5999]

I don't fully understand everything up there, but at this point I realize it's probably because I haven't done enough reading on this specific topic. I'll go through some more material and try to understand what you've said. Thanks for your help, Dick!

 

[Dick]

Very welcome, lonewolf5999.