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This question has been bothering me for a while now. Everywhere I look, everyone talks about 'fictitious forces' and how they apparently explain the bike being in equilibrium. However, if we just look at a simple force diagram, we can see that turning moments around certain points are unbalanced:Force diagram

If we take moments around, for example, the point where the wheel touches the ground, we obtain that a resultant moment of $amg\cos \theta$, where $a$ is some length, $m$ the mass of the combined system of the driver and the bike, and $\theta$ the angle between the bike and the ground, is acting so as to make the system fall down in the direction of the centre.

This moment will always be 'towards the centre' - irrespective of the bike's position. That means it should fall down. So why doesn't it? Clearly, either there should be no resultant moment, or the resultant moment throughout the duration of one lap made by the bike should be $0$, but none is evidently the case here. What am I missing?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/506/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Feb 22, 2018 at 17:37
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    $\begingroup$ Strictly speaking, you are using statics analysis of torques to an accelerating axis of rotation. Not appropriate. $\endgroup$
    – npojo
    Commented Feb 22, 2018 at 18:44
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    $\begingroup$ @npojo Okay, so tell me how it's not appropriate, then? Instantaneously, there is a torque acting towards the centre at any given moment. Why does the torque have no effect? $\endgroup$
    – Max
    Commented Feb 22, 2018 at 18:57
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    $\begingroup$ @Qmechanic I think I am asking a different question. What you linked is talking about a bicycle at high speed - I am talking about a bicycle going in circular motion. $\endgroup$
    – Max
    Commented Feb 22, 2018 at 19:09
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    $\begingroup$ @Pieter Ok, I think I see it now. I forgot about the moment of inertia. At one side of the circle that the bike makes, the torque is in the opposite direction to the torque at the opposite side, the latter being carried all the way to the first side via inertia. The two moments cancel out. In other words, the integral with respect to time between 0 and the time it takes for the bike to get from one side to another of the torque acting on the bike over mass should be greater than 0. $\endgroup$
    – Max
    Commented Feb 22, 2018 at 20:12

4 Answers 4

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In the inertial frame, the torques are unbalanced and the rider's rotational momentum about a point on the ground changes. It's just that this change does not result in the rider toppling.

If instead you consider the frame where the rider is at rest, then this (accelerating) frame will have fictitious forces opposite the acceleration appear.

These forces will act through the center of mass of the rider, and will be in the opposite direction of the acceleration. Since the rider is accelerating to the left, there should be a fictitious frame force to the right. This balances the torques about the tire contact.

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  • $\begingroup$ Yep. Rotational momentum. I didn't think about it. $\endgroup$
    – Max
    Commented Feb 22, 2018 at 20:14
  • $\begingroup$ @Max -You do not even need to consider a "rotational momentum". It is just a static cancellation of torques in the rotating frame. See the quantitative description in my extended answer. $\endgroup$
    – freecharly
    Commented Feb 22, 2018 at 22:41
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The reason that the biker doesn't fall when leaning in a curve is the centrifugal force on the CM of the biker. You see this best when the biker moves in a circle in a curve. The torque by the gravitational force is exactly balanced by the torque due to the centrifugal force on the biker. A description you will find here.

Quantitative description: As you seem not to accept that the centrifugal force explains the problem, I give you a quantitative derivation. The torque $T_g$ of gravitational force on the center of mass relative to the intersection of axis driver/bike with the ground is $$T_g=mgacos\theta$$ where $a$ is the distance of the center of mass from the intersection of the axis with the ground. The opposing torque of the centrifugal force on the center of mass with respect the same point on the ground in the frame rotating with the biker is $$T_{cf}=\frac{amv^2 sin \theta}{r}$$ where $v$ is the velocity of the bikers center of mass and $r$ is the (local) radius of the bikers center of mass path. Thus you have a resultant torque zero for $$T_g=mgacos\theta=\frac{amv^2 sin \theta}{r}=T_{cf}$$ From which follows the inclination of the axis of the biker in equilibrium $$tan \theta=\frac {g r}{v^2}$$ This shows that the resulting torque acting on the biker is indeed zero for a certain angle of inclination, which confirms the experience of all bikers driving in a curve.

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    $\begingroup$ You are not answering my question. I know that all the moments do indeed cancel out in real life, as we don't fall while leaning our bike. My question is why. Taking moments around the point where the wheel touches the ground, torques are unbalanced. Centrifugal force has nothing to do with my question, which was actually the first thing I mentioned in op. $\endgroup$
    – Max
    Commented Feb 22, 2018 at 19:03
  • $\begingroup$ @Max -Perhaps you should add this supposed unbalancing of torques to your diagram so that everybody can understand what your problem is. The diagram doesn't show the centrifugal force at all. Maybe you take a look at the link I gave in my answer. $\endgroup$
    – freecharly
    Commented Feb 22, 2018 at 19:12
  • $\begingroup$ yeah, that seems like a good idea; I'm on my phone now, so I can't do that now, but as soon as I'm on my computer, I'll do that. However, I thought I'd explained the imbalance clearly in the op? Take moments around the point where the wheel touches the ground, and the only torque acting is the one due to gravity. Centrifugal force is not on the diagram because it is not a real force. I did click the link you provided, but it doesn't really answer my question. I am talking about the unbalancing of torques around the point where the wheel touches the ground. $\endgroup$
    – Max
    Commented Feb 22, 2018 at 19:30
  • $\begingroup$ Also, you mentioned the 'supposed unbalancing of torques'. Can I ask what exactly is not clear in my question (not being defensive, just an honest good-faith question) $\endgroup$
    – Max
    Commented Feb 22, 2018 at 19:33
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    $\begingroup$ I agree with @freecharly, the apparent imbalance is because of OP's failing to show the torque moment created by centrifugal force. As it is the question is not complete. Otherwise once we consider the centrifugal force, the biker would fall out if he/she doesn't lean in to the turn. Stunt car drivers take advantage of centrifugal force to ride the car on two wheels on one side by initiating a tight turn and lifting one side up! If a pilot doesn't bank in to the turn the plane will roll out and stall. $\endgroup$
    – kamran
    Commented Feb 22, 2018 at 23:50
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Indeed if the bike was travelling in a straight line, the bike would fall over. It is the circular motion that has the rider falling towards the centre of the circle, but the constant change in direction prevents the bike from simply falling to the ground.

Think of a velodrome where bike speed races are held. When the rider is moving along the banked curve, what would happen if the curve became straight but still banked? Well the rider would begin to drop down the bank, not because he is forcing the bike there, but because this would naturally occur. His bike would be falling over due to the imbalance of the moments you mention.

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  • $\begingroup$ How does constant change in direction prevent the bike from falling? The resultant torque is still acting towards the centre. The bike should fall even while constantly changing direction. $\endgroup$
    – Max
    Commented Feb 22, 2018 at 19:05
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Simply because in doing the calculations in the frame of reference of the bike itself, you are doing these calculations in an accelerated frame of reference, and you need to factor in the acceleration to balance everything.

If you did the calculations in an inertial frame of reference, then, in that frame, the moment does exist, but it is balanced by the moment created by the lateral acceleration through the center of mas to the outside of the turn.

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