# Alternative approach -- Bicycle on a curve

Tags:
1. Jul 28, 2017

1. The problem statement, all variables and given/known data

A bicycle of mass m is travelling at constant speed v around a curve of radius r without slipping. You can take the acceleration due to gravity as g. Calculate the angle of tilt, θ, that will enable it to balance.
2. Relevant equations
R=mg
Rsintheeta (Length)=Fcostheeta (Length); R is the normal force, F is the centripetal force and Length is the distance from the ground to the center pf mass
F=mv^2/r
3. The attempt at a solution
I solved this question by taking moments about the center of mass and using the above equations, and a bit of substitution. However, I have two queries:
i) Friction provides centripetal force and acts towards center (Sliding Friction. Right?). But friction should also act opposite to the direction of motion (tangent along the entire path). Why don't we account for it in the such problems. It is called rolling friction, I guess. Correct me if I am wrong.
ii) Secondly, if I take moments about the point of contact, instead of at the center of mass, the equation doesn't seem to balance as there's moment due to weight on one side of the equation and all the other moments turn out to be zero since their line of action passes through the pivot.

2. Jul 28, 2017

### haruspex

No. The tyres are in rolling contact, so static friction.
Kinetic friction acts to oppose actual relative motion of surfaces in contact.
Static friction acts to oppose that relative acceleration that would occur were it not present.
Without friction, the bicycle would accelerate out radially relative to its actual path.
The forward motion of the mass centre imbues it with angular momentum about the instantaneous point of contact. As a vector, that points radially. The torque generated by mg results in precession. In terms of the linear motion, that means the bicycle goes around the circle.

3. Jul 28, 2017

### CWatters

The tyre isn't sliding or trying to slide in that direction so there is no friction force in that direction. There might be air resistance or rolling resistance but that's not mentioned in the problem so can be ignored.

4. Jul 28, 2017

### Staff: Mentor

When you take moments about an axis that is not through the center of mass in an accelerating system, you need to include the moment of the pseudo force through the center of mass.

5. Jul 28, 2017

### haruspex

Asad Raza took moments about the "point of contact". That is ambiguous because it is not clear whether it is the point on the road or a (moving) point on the tyre, but I took it to be the former. That makes it an inertial frame, so no pseudo forces.
Did you read my explanation in post #2?

6. Jul 28, 2017

### Staff: Mentor

The above figure shows the forces acting on the bike with- and without the pseudo force shown. To do the moment balance correctly in an approach where the pseudo force is not included (in a problem involving accelerations), one must take moments about the center of mass. To do the moment balance correctly in an approach where the pseudo force is included, one can take moments about any point.

In either case, if moments are taken about the center of mass, the moment balance is given by:
$$RL\cos{\theta}=mgLcos{\theta}=\frac{mv^2}{R}L\sin{\theta}$$
This is the moment balance obtained by Asad Raza.

But when Asad Raza took moments about point A in the diagram to the left (without the pseudo force included), he obtained:

$$mgL\cos{\theta}=0$$This is, of course, incorrect.

But, if we take moments about point A in the diagram on the right (with the pseudo force included), we obtain:
$$mgL\cos{\theta}=\frac{mv^2}{R}L\sin{\theta}$$
This is in correct, and in agreement with the moment balance about the center of mass.

The key learning here is that, in a system that is not in equilibrium (i.e., involving acceleration), if you include the pseudo force, it doesn't matter which point you choose to take moments about. Any point other than the CM or point A would deliver the same result. But, if the pseudo force is not included, it is only valid to take moments about the CM.

7. Jul 28, 2017

### haruspex

Seems Chet and I disagree. Private conversation initiated.

8. Jul 30, 2017

### haruspex

As a result of that private thread, I have developed all the equations for the point-of-contact reference frame, so I might as well post them here.

With the point of contact as origin, the centre of mass is at some vector $\vec c$. That means it has angular momentum $m\vec c\times\vec v$. The weight exerts torque $m\vec c\times\vec g$. This leads to a precession about a vertical axis.
If the precession rate is $\vec\omega$, the torque is $\vec \tau$ and the angular momentum is $\vec L$ then $\vec\tau=\vec\omega\times\vec L$.
Thus $\vec c\times\vec g=\vec \omega\times(\vec c\times\vec v) = \vec c(\vec\omega.\vec v)-\vec v(\vec\omega.\vec c)$.

Since $\vec\omega$ and $\vec v$ are orthogonal:
$\vec c\times\vec g=-\vec v(\vec\omega.\vec c)$
This is reasonable since $\vec v$ is indeed orthogonal to both $\vec g$ and $\vec c$.

The situation is analogous to force and linear momentum. To the extent that an applied force is parallel to the existing momentum it leads to a change in speed; to the extent that it is normal to the momentum it leads to a rate of change of direction (slew rate - I think that's the right word).

Similary, to the extent that a torque is parallel to the angular momentum it leads to a change in its magnitude; to the extent that it is normal it leads to a precession rate, i.e. the rate of change of direction of the angular momentum.
In the present case, the torque is normal to the angular momentum.

Using c for $|\vec c|$ etc., theta for the tilt of the bicycle, R for the horizontal distance from the mass centre to the centre of the circle:
$\vec v =\vec \omega\times\vec R$
$v=\omega R$
$|\vec c\times \vec g|=cg\sin(\theta)$
$|\vec v(\vec\omega.\vec c)|=v\omega c\cos(\theta)$
Combining all those equations we find $v^2=Rg\tan(\theta)$

9. Jul 31, 2017

### Staff: Mentor

Hi haruspex,

As I said in our private conversation, I was not able to follow what you have done. Any chance of providing a diagram.

One question, regarding $\theta$, that is the angle of tilt of the bike with respect to the horizontal in your analysis, correct?

10. Jul 31, 2017

### haruspex

No, to the vertical.

11. Jul 31, 2017

### Staff: Mentor

One more question. Is what I did incorrect? If so, why was I able to get the same answer as you?

12. Jul 31, 2017

### haruspex

You effectively used a non-inertial frame. To ignore the precession, your point of contact axis must be moving with the bicycle. Thus the pseudo-force was necessary.

13. Jul 31, 2017

### Staff: Mentor

Is that an incorrect thing to do?

14. Jul 31, 2017

### haruspex

Not at all, but as I wrote in post #5 it is not entirely clear whether @Asad Raza was taking the "point of contact" frame of reference as the instantaneous point on the ground (i.e. a fixed point, so an inertial frame), or as the moving point of contact. And, as I wrote there, I assumed Asad was taking the former view and could not understand how there could be an unbalanced torque. In that view, the explanation is that there is an unbalanced torque, but the consequence of it is precession, not a change in tilt of the bicycle.
If one takes the moving point of contact as axis then there is a pseudo force to be included, which is the way you approached it.

As far as right or wrong is concerned in the context of this thread, it depends on which view Asad intended. If Asad intended the inertial frame then your approach, while valid in itself, did not address the question posed.