# Bicycle problem

1. Dec 9, 2007

### olga11

[SOLVED] Bicycle problem

Will the bicycle move forward or backward?

We have 2 forces. The friction and the one we act on the wheel through the string. The weight and the normal force have torque equal to zero.

torque of the friction = T.r1
torque of the F = F.r2

In the second picture r1>r2, so the wheel will move backward, since in order for the wheel to move F=static friction

How am I doing so far? Am I in the right path, please?

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2. Dec 10, 2007

### catkin

Interesting question!

Assumptions:
• the tricycle is initially at rest.
• friction is negligible except between wheel and floor.
• force in string is not sufficient to cause slippage between wheel and floor (could also analyse case where this is not true).
• torques about vertical axes are negligible.

Let:
• R be the radius of the wheel.
• r be the distance from the wheel axle to the pedal axle.

For the first picture I, too, think there are 4 forces acting on the wheel but they are not the same as your 4:
1. Weight of wheel itself acting vertically downward through axle and wheel/floor contact point.
2. Contact force from floor.
3. Tension in string acting horizontally, line of action R + r above floor.
4. Force of axle acting on wheel, line of action unknown.

Must sleep now. It's the 4th force that's interesting; if the tricycle is accelerating forward (I'm sure it is, it's the analysis that's interesting) then it includes a horizontal component that is accelerating the mass of the tricycle and rotationally accelerating the rear wheels).

Last edited: Dec 10, 2007
3. Dec 11, 2007

### catkin

A further iteration (this is exploratory) ...

Assumptions:
• the tricycle is initially at rest
• friction is negligible except between wheel and floor
• all wheels are identical
• tension in string is small?
Let:
• R be the radius of the wheel.
• r be the distance from the wheel axle to the pedal axle.
• I be the moment of inertia of each wheel
• m be the mass of each wheel
• M be the mass of tricycle without wheels
For the first picture, the forces acting on the wheel:
1. Weight of wheel itself acting vertically downward through axle and wheel/floor contact point.
2. Contact force from floor.
3. Tension in string acting horizontally, line of action R + r above floor.
4. Force of axle acting on wheel, line of action unknown.
Resolving horizontally:
1. No horizontal component.
2. Friction acting horizontally, line of action at floor level.
3. Tension in string acting horizontally, line of action R + r above floor.
4. Horizontal component of axle acting on wheel.
Resolving vertically:
1. Weight of wheel itself acting vertically downward through axle and wheel/floor contact point.
2. Normal component
3. No vertical component.
4. Vertical component of axle acting on wheel.
Perhaps it is useful to consider the wheel as a lever hinged at its contact point with the floor. When the string has moved a small distance ${\delta}s$ in short time ${\delta}t$ it has pulled the wheel axle forward ${{\delta}s R} / {(R + r)}$ and rotated the wheel $sin^{-1}({\delta}s / r)$ (the latter using small angle approximation).

During the short time ${\delta}t$ the accelerations (translational and rotational) from rest are constant so the final velocity is twice the average velocity.

Final translational velocity of wheel axle is
$$2 \times \frac{{\delta}s R}{{\delta}t(R + r)}$$

Translational acceleration of axle (${\Delta v} / {\Delta t}$)
$$2 \times \frac{{\delta}s R}{{\delta}t^2(R + r)}$$

Similarly the final rotational velocity and the rotational acceleration are
$$2 \times \frac{sin^{-1}({\delta}s / r)}{\delta t}$$
and
$$2 \times \frac{sin^{-1}({\delta}s / r)}{\delta t^2}$$

4. Dec 11, 2007

### catkin

A further iteration (this is exploratory) ...

Assumptions:
• the tricycle is initially at rest
• friction is negligible except between wheel and floor
• all wheels are identical
• tension in string is small?
Let:
• R be the radius of the wheel.
• r be the distance from the wheel axle to the pedal axle.
• I be the moment of inertia of each wheel
• m be the mass of each wheel
• M be the mass of tricycle without wheels
For the first picture, the forces acting on the wheel:
1. Weight of wheel itself acting vertically downward through axle and wheel/floor contact point.
2. Contact force from floor.
3. Tension in string acting horizontally, line of action R + r above floor.
4. Force of axle acting on wheel, line of action unknown.
Resolving horizontally:
1. No horizontal component.
2. Friction acting horizontally, line of action at floor level.
3. Tension in string acting horizontally, line of action R + r above floor.
4. Horizontal component of axle acting on wheel.
Resolving vertically:
1. Weight of wheel itself acting vertically downward through axle and wheel/floor contact point.
2. Normal component
3. No vertical component.
4. Vertical component of axle acting on wheel.
Perhaps it is useful to consider the wheel as a lever hinged at its contact point with the floor. When the string has moved a small distance ${\delta}s$ in short time ${\delta}t$ it has pulled the wheel axle forward ${{\delta}s R} / {(R + r)}$ and rotated the wheel $sin^{-1}({\delta}s / r)$ (the latter using small angle approximation).

During the short time ${\delta}t$ the accelerations (translational and rotational) from rest are constant so the final velocity is twice the average velocity.

Final translational velocity of wheel axle is
$$2 \times \frac{{\delta}s R}{{\delta}t(R + r)}$$

Translational acceleration of axle (${\Delta v} / {\Delta t}$)
$$2 \times \frac{{\delta}s R}{{\delta}t^2(R + r)}$$

Similarly the final rotational velocity and the rotational acceleration are
$$2 \times \frac{sin^{-1}({\delta}s / r)}{\delta t}$$
and
$$2 \times \frac{sin^{-1}({\delta}s / r)}{\delta t^2}$$

Mmm ... that can't be right; in the limit, as ${\delta}t$ approaches zero, the accelerations become very small.

5. Dec 11, 2007

### olga11

Can we ignore friction between the wheel and the floor?

6. Dec 12, 2007

### catkin

We can choose one of three frictions:
1. none
2. not enough (there is some friction but the wheel slides on the floor)
3. enough (the wheel does not slide on the floor)
The first is unrealistic. The second is when the tension in the string is high and and hence accelerations high; this is when the string is jerked; the position of the hand in the pictures does not look as if it is jerking the string. That leaves the third as the most appropriate choice.

7. Dec 12, 2007

### olga11

For the first picture how will we justify that the wheel is moving forward?

Won't we say that this happens because the torque of the tension is bigger?

For the second picture: Initially the wheel is moving backwards because the torque of the static friction is bigger. Then what?

Last edited: Dec 12, 2007
8. Dec 12, 2007

### Shooting Star

9. Dec 12, 2007

### olga11

I believe F in the figure is the force we act on the yo-yo, not the friction.

10. Dec 12, 2007

### Shooting Star

That's true, but I changed it to P and called the frictional force F.

The thing to understand is, always the wheel or the cycle will tend to move in the direction of the applied force.

11. Dec 12, 2007

### olga11

So the wheel will move towards P in any case (transportational movement)

In my first picture it will rotate anticlockwise all the time.

In my second picture it will rotate clockwise at the beginning and then?

12. Dec 12, 2007

### Shooting Star

As the wheel rotates clockwise, the point of application of the force will shift (since you are pulling with a string tied to a fixed point on the wheel), until the torques balance so that there is no tendency of the wheel to rotate, and it will be simply dragged along.

The same thing will happen eventually in the first case too, because the point of application of the force will shift in both cases. Anyway, we are considering what will happen right at the beginning.

13. Dec 13, 2007

### olga11

If the radius of the pedal is extended to be equal to the radius of the wheel or longer, what will it happen?

Torque(pedal) = Force(acted by the cyclist) * Radius (pedal-arm)

A longer pedal means more torque, so the bicycle accelerates faster for the same force provided by the cyclist.

If the velocity is constant, the cyclist can provide less force to achieve the same speed.

I need help to connect the radius of the wheel to the radius of the pedal.

14. Dec 13, 2007

### Shooting Star

(Isn't it a tricycle in the pictures?)

If you make the pedal arm longer than the radius of the wheel, the pedal will touch the ground! Of course, then you can get off, lift the cycle, move the pedal and then start to pedal until the pedal touches ...

But if we just analyze the motion for the part of the revolution when the pedal doesn't touch the ground, the end of the pedal traverses more distance. The cyclist may provide less force, but has to do it for a longer distance and so work done would be the same, but the power output would be less, which would be convenient. Modern bicycles, where the pedal wheel and the rear wheel are connected by a chain, are more or less optimized for such a scenario. But for the tricycle, from a practical point of view, it'll be difficult to push the pedal when it'll extend so far from the seat.

There has been many weird concepts in the history of the bicycle, and the tricycle too. Look up http://en.wikipedia.org/wiki/Penny-farthing and http://en.wikipedia.org/wiki/Tricycle.

15. Dec 13, 2007

### olga11

I liked it very much and I copied it in my answer. I hope my teacher has the same sense of humor.

16. Dec 13, 2007

### YellowTaxi

If the trike didn't exist (just the front wheel on it's own)
then the front wheel would definitely roll to the left.
because you would just be applying a torque from about 5cm above ground level
no different from pushing or pulling it form any other place really
- it moves left

If the trike is attached but completely frictionless
Then it only adds mass (inertia) to the centre of the front wheel

So, yes it always moves to the left
regardless of how much friction in the front tyre
- it can slip to the left if it pleases, I don't care.

(I think)

lol