# Bicycle wheel rotational motion

1. Oct 29, 2015

### Haveagoodday

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

A guy is riding a bicycle, and we consider the front wheel, which has mass

M, radius R and for the purpose of the moment of inertia can be thought of

as a uniform disc.

a) When the bike is going with linear speed v, what is the magnitude and

direction of the angular momentum of the wheel, around its centre of mass?

b) At some instant, what is the angular momentum of the wheel around the

point at which it touches the ground?

c) The bike guy now comes slightly off balance, so that the front wheel is

not completely upright, but leans over with an angle from vertical. What

is the magnitude and direction of the torque from gravity, around the point

where the wheel touches the ground?

d) Which the way does the wheel try to turn as a result of this torque?

3. The attempt at a solution
Can somebody check on my solutions, that would be much appreciated
a)
ω=v/r
I=1/2MR^2
L=Iω=1/2MRv

θ=arcsin MRv/2L

b) L=Iω=3/2MRv

c) τsinθ=3/2MgRcosθ
θ=arcsin (3/2MgRcosθ)/ τ

d) to the left

Last edited by a moderator: Oct 29, 2015
2. Oct 29, 2015

### haruspex

Yes.
I have no idea what makes you think that's the direction of the angular momentum. When an object rotates about an axis, what direction is its angular momentum vector in?
Yes.
You don't have to consider the wheel's rotation here. Just consider a wheel leaning at angle theta to the vertical. What are the forces on it? What torque results?
The question does not say whether the bicycle leans to left or right.

3. Nov 2, 2015

### Unikumet

May I ask, how did you get L=Iw=3/2 MVr in task b)?

4. Nov 2, 2015

### J Hann

I agree with your answers for parts (a) and (b) (using the parallel axis theorem for part (b)).
The angular momentum vector for the spinning wheel is of course in the horizontal plane.
Torque = R X F the vector cross product which is also in the horizontal plane.
Thus the change in angular momentum must also be in the (nearly horizontal) plane since
the change in angular momentum is in the direction of the torque vector.
Note that when a cyclist starts to tip that the cyclist will turn the front wheel to increase the
centrifugal force due to the decreasing radius of the turn to again right the bicycle.
Also, there is a "caster" effect which tends to straighten the path.
This is like the caster wheels on furniture which tend to straighten themselves in the
direction in which you are trying to move the furniture.

5. Nov 2, 2015

### haruspex

To clarify, none of that is relevant to answering this question, right?
(I dislike the term "caster effect" in this context. the two processes are somewhat different.)

6. Nov 3, 2015

### J Hann

The comments are irrelevant to the question, as asked, as long as one does not take away the impression
that the angular momentum of the wheels are the main factor in considering the stability of a bicycle.

7. Nov 3, 2015

### haruspex

Understood.
What makes bicycles rideable is a much researched topic, and the answer appears quite complex. But the question in the OP does not go as far as considering how the gravitational torque affects the steering (it just gives one the feeling that is where it is headed).

8. Nov 4, 2015

### jensjensen

In a), wouldn't the answer be 0, since we are looking at the center of mass?

9. Nov 4, 2015

### haruspex

No, why would that follow? The wheel is rotating. Angular momentum about an axis is $I\omega$, where I is the moment of inertia about the axis and $\omega$ is the rotation rate.

10. Nov 4, 2015

### rcgldr

a) Why include θ in the answer? Using the shown diagram, using right hand rule, the direction of angular momentum is towards the viewer (perpendicular to the image).

c) torque = radius x force. If θ is angle from vertical, then gravitational force perpendicular to wheel is = m g sin(θ). The question asks for the torque about the point of contact, so any forces related to the ground are ignored.

d) angular momentum is towards the viewer, torque due to gravity is to the right if the bike leans towards the viewer, to the left if the blike leans away from the viewer. Angular velocity vector due to precession is up if the bike leans towards the viewer, down if the bike leans away from the viewer. The front edge of the wheel would steer into the same direction as the lean.

The trail / caster as depicted in the image would also steer the front wheel into the direction of lean, but that is not being questioned in this problem.

Last edited: Nov 5, 2015
11. Nov 4, 2015

### haruspex

Unfortunately @Haveagoodday seems to have lost interest. He/she has not responded to any of the posts. Maybe having too good a day.

12. Nov 4, 2015

### rcgldr

13. Nov 5, 2015

### StavangerFinest

It absolutely does and than you for that. I'm sure that this is helpful to a lot of people like myself, so never quit just if the initiator of the question doesn't seem to response.
Thanks again, and @haruspex I'm still waiting on response on one of my previous questions abt angular velocity exercise, help would be appreciated

14. Nov 5, 2015

### sludgefactory

Even though haveagoodday left, you have no idea how many students you guys are helping! thanks a lot!

15. Nov 5, 2015

### Haveagoodday

Does that mean the torque of gravity is 0? Thank you for helping!

16. Nov 5, 2015

### haruspex

No.
Draw a free body diagram of the wheel as seen from the rear. The wheel is leaning at angle theta to the vertical. Show the forces acting on it.
Where does the force of gravity act? What is its moment about the point of contact of the wheel with the ground?