Why does the Motorcycle does A Wheelie when you Clutch-Wheelie

[jbriggs444]

I think the wheelie can be simply explained through these 2 horizontal forces (at the contact patch and at the center of mass). However, some explanations I have seen also consider the torque of the engine on the rear wheel, and an opposing torque that lifts the front end as has also been posted in this thread here

Explanations involving torque of the wheel or the tension in the chain should not be thought of as competing explanations. They are simply different ways of looking at the same physical situation.

In my thinking this [chain tension as an explanation] is not true. The torque on the engine creates chain tension which turns the rear wheel. This chain tension is a horizontal force between the rear wheel sprocket and the engine sprocket. It is an internal force and creates no torque about the COM.

Whether the chain tension is an "internal" or an "external" force depends on where you draw imaginary lines around the system of interest. If you concentrate on the rigidly connected bicycle frame, engine and rider then the rear wheel is not part of the system. (And if you concentrate on the rear wheel, the bicycle frame, engine and rider are not part of the system). Either way chain tension is an external force.

But chain tension is not the only external force. If all you had was chain tension, the rear wheel would be drawn forward to the engine. That does not happen because the rear wheel is mounted on an axle that is attached to the frame. So you also have frame compression which amounts to a second external force between wheel and frame/engine/rider. Taken together, these two forces are a "couple" -- a pair of equal and opposite forces that amount to a pure torque.

 

[Dr. Headholio]

There is no contradiction here, just different ways to divide the system into bodies, that determines which forces/torques are external/internal. If you model the back-wheel and rest-bike as two separate bodies, then it's the torque from the chain force that rotates the rest-bike. If you model the entire bike as one body, then it's the torque from the ground reaction.

Thanks for the explanation, but I'm having trouble seeing it. As I mentioned, I view the chain force as a horizontal internal force that can not lift the front.
To simplify, I imagine a bicycle with the rear wheel strapped to the ground. Pushing on the pedal will create chain tension, trying to pull the rear wheel axle toward the crank. The reaction to the chain tension is an internal opposite force in the frame. the reaction to the force on the wheel/ground attachment is an opposite force in the ground. There is no moment created that I can see.
If the rear wheel is cut loose, then the COM is accelerated and the torque is F x r = Ma x r and is the distance from the COM to the point of rotation.

Or, is chain tension not an internal force, and it pulls on the front of the bike and causes a moment about the contact patch?

 

[Dr. Headholio]

Explanations involving torque of the wheel or the tension in the chain should not be thought of as competing explanations. They are simply different ways of looking at the same physical situation.Whether the chain tension is an "internal" or an "external" force depends on where you draw imaginary lines around the system of interest. If you concentrate on the rigidly connected bicycle frame, engine and rider then the rear wheel is not part of the system. (And if you concentrate on the rear wheel, the bicycle frame, engine and rider are not part of the system). Either way chain tension is an external force.

But chain tension is not the only external force. If all you had was chain tension, the rear wheel would be drawn forward to the engine. That does not happen because the rear wheel is mounted on an axle that is attached to the frame. So you also have frame compression which amounts to a second external force between wheel and frame/engine/rider. Taken together, these two forces are a "couple" -- a pair of equal and opposite forces that amount to a pure torque.

I posted above before I saw your post. I guess I'm going to have to think about this some more.

 

[A.T.]

...I view the chain force as a horizontal internal force ... is chain tension not an internal force...

Did you even read what I wrote? Which forces are internal/external depends on how you define your bodies.

 

[Dr. Headholio]

There is no contradiction here, just different ways to divide the system into bodies, that determines which forces/torques are external/internal. If you model the back-wheel and rest-bike as two separate bodies, then it's the torque from the chain force that rotates the rest-bike. If you model the entire bike as one body, then it's the torque from the ground reaction.

I can see that chain tension can create a force on the frame that is pointed toward and above the rear axle, thereby creating a moment about the rear axle.

But I don't think this explains a wheelie, nor is it even necessary to have this force to perform a wheelie. For example, imagine a motorcycle engine with a drive shaft that goes to the rear wheel , and a gear box converts the torque of the spinning drive shaft to forward spinning of the rear wheel. In this case, I can see no opposite torque that lifts the front wheel. If the rear wheel were nailed to the ground, the torque of the engine would try to twist the engine in the opposite direction but not create any upward force on the frame that would lift the front wheel.

My point is that the arguments that say that the torque on the wheel which spins it forward creates an opposite torque on the frame to spin it the other direction is not true. The wheel lifts from the friction force on the ground, and the acceleration of the center of mass. If there is a chain force, it can add to the forces lifting the wheel, but there can be torque on the rear wheel without an opposite torque lifting the frame, as in the case of a driveshaft driven rear wheel. Am I thinking about this correctly?

 

[davenn]

For example, imagine a motorcycle engine with a drive shaft that goes to the rear wheel , and a gear box converts the torque of the spinning drive shaft to forward spinning of the rear wheel. In this case, I can see no opposite torque that lifts the front wheel. If the rear wheel were nailed to the ground, the torque of the engine would try to twist the engine in the opposite direction but not create any upward force on the frame that would lift the front wheel.

My point is that the arguments that say that the torque on the wheel which spins it forward creates an opposite torque on the frame to spin it the other direction is not true.

why do you believe that ?

forget about where the power to rotate the wheel is coming from. think about the forces between the wheel and axle turning one way and the axle housing ( the diff) and the rest of the body of the car/bike wanting to move in the opposite direction with equal and opposite force

same as what is happening in a helicopter the driveshaft and main rotors rotating one way and the rest of the body of the helicopter wanting to rotate the other way.
without the tail rotor ( or exhaust blast on some) the body will freely spin in the opposite directionD

 

[A.T.]

For example, imagine a motorcycle engine with a drive shaft that goes to the rear wheel , and a gear box converts the torque of the spinning drive shaft to forward spinning of the rear wheel. In this case, I can see no opposite torque that lifts the front wheel.

If the rest-bike applies a torque to the rear wheel, then the rear wheel applies an equal but opposite torque to the rest-bike. See third Law here:

http://www.4physics.com/phy_demo/Newton/Newton_rot2.htm

 

[sophiecentaur]

If the rear wheel were nailed to the ground, the torque of the engine would try to twist the engine in the opposite direction but not create any upward force on the frame that would lift the front wheel.

That just has to be nonsense. The bike does not 'know' whether the wheel is nailed down or just has good grip on the road. The torques 'internal' to the bike are all balanced out because the rotating parts of engine and transmission are contained in bearings and housings, which provide restraints. The only unbalanced torque is the torque that turns the wheel (ignoring the MI of the working parts). If you nail the wheel to the ground, this torque will tend to cause the bike to rotate (lift the front wheels). That's a bit simplistic because it is always necessary to use the clutch as the engine has no torque at zero revs and a certain amount of rotational momentum can also help in the initial lift. Also, the engine may even operate at a more suitable speed is there is some wheel spin. Nevertheless, it all comes down to a torque about the rear spindle which is in a sense that will tend to lift the front wheel.

 

[A.T.]

For example, imagine a motorcycle engine with a drive shaft that goes to the rear wheel , and a gear box converts the torque of the spinning drive shaft to forward spinning of the rear wheel. In this case, I can see no opposite torque that lifts the front wheel.

 

[sophiecentaur]

The wheel lifts from the friction force on the ground, and the acceleration of the center of mass. If there is a chain force, it can add to the forces lifting the wheel, but there can be torque on the rear wheel without an opposite torque lifting the frame, as in the case of a driveshaft driven rear wheel.

Try this: Imagine, instead of letting the clutch in, you have the back brake on and someone pulls the road sharply forward. The force on the wheel periphery will be applied to the wheel and, hence, to the spindle / axle. The reaction force (due to the mass of the bike / car) will produce a couple which will tend to lift the front wheels off the ground. If the friction is enough to stop slipping then the vehicle will be moved forwards and, if the acceleration is great enough, the car will actually tip up. Try it with a book , on its edge, or a box on a table. Push horizontally against the bottom corner, next to the table and you can make the book / box rotate, like a wheelie. The ONLY difference between this and the wheelie on a bike is that the force is driven from the wheel and not the road, accelerating. There is a torque in both cases.

 

[Dr. Headholio]

If the rest-bike applies a torque to the rear wheel, then the rear wheel applies an equal but opposite torque to the rest-bike. See third Law here:

http://www.4physics.com/phy_demo/Newton/Newton_rot2.htm

"For every applied torque, there is an equal and opposite reaction torque"

Does this also apply to moments created by a force vector?

Imagine a spool of string able to freely rotate about an axle in space. Tie a rock to the end of the string and drop it. The rock applies a force to the string, the string applies a torque to the spool, the spool applies an equal and opposite force to the string, but is there an equal and opposite torque?

 

[sophiecentaur]

but is there an equal and opposite torque

There is an equal and opposite force on the string so there must be an equal and opposite torque on the drum. That doesn't mean the drum is not accelerating. In the same way, an object accelerates due to tension in the string that pulls it. Do no confuse Newton's Third law with 'Equilibrium'. They are two different things.

 

[Dr. Headholio]

There is an equal and opposite force on the string so there must be an equal and opposite torque on the drum. That doesn't mean the drum is not accelerating. In the same way, an object accelerates due to tension in the string that pulls it. Do no confuse Newton's Third law with 'Equilibrium'. They are two different things.

But my point is that the "equal and opposite reactions" are the string pulling on the spool, and the spool pulling on the string. The reactions are forces. One of the forces creates a torque, and the other does not. I don't see equal and opposite torques in this case.
I can envision the equal and opposite torques idea for the helicpoter example, or a hand turning a screwdriver. Is there something specific about the way that torque must be defined that makes it different than a moment created by a force?

 

[Dr. Headholio]

Here is a another example, trying to show that a "torque reaction" is not necessary to explain a wheelie.

You build a sidewalk to connect 2 very tall buildings, and put a bicycle on the sidewalk. There is a spool mounted above the sidewalk that holds a very long bicycle chain. You attach a weight to the end of the chain, engage the chain on the rear wheel sprocket, and throw the weight off the sidewalk.
The weight falls and pulls the chain, which turns the rear wheel and propels the bike forward.

The only attachment between the wheel and the frame is the frictionless hub bearing. No torque can be transferred from the wheel to the frame (only translational force).

There is a torque on the wheel (mass of the weight x radius of the sprocket), and the equal and opposite reaction is the sprocket pulling on the chain. No opposite torque that I can see.

Can you have wheelie in this situation? Yes! The heavier the weight, the faster the bike accelerates. Eventually the center of mass of the bike plus rider is accelerated to the point where there is a force at the center of mass creating a moment about the rear wheel which will overcome the opposite moment of the mass holding the front wheel down. .

 

[A.T.]

Imagine a spool of string able to freely rotate about an axle in space. Tie a rock to the end of the string and drop it. The rock applies a force to the string, the string applies a torque to the spool,...

...around the center of the spool.

the spool applies an equal and opposite force to the string, but is there an equal and opposite torque?

Yes, around the center of the spool.

 

[A.T.]

Here is a another example,

Do you understand why your first example was wrong?
https://www.physicsforums.com/threa...you-clutch-wheelie.687715/page-2#post-5148039

...trying to show that a "torque reaction" is not necessary to explain a wheelie.

Your example is not how a bike or car is propelled and thus doesn't show what is necessary to explain a wheelie on an actual bike or car.

You build a sidewalk to connect 2 very tall buildings, and put a bicycle on the sidewalk. There is a spool mounted above the sidewalk that holds a very long bicycle chain. You attach a weight to the end of the chain, engage the chain on the rear wheel sprocket, and throw the weight off the sidewalk.
The weight falls and pulls the chain, which turns the rear wheel and propels the bike forward.

The only attachment between the wheel and the frame is the frictionless hub bearing. No torque can be transferred from the wheel to the frame (only translational force).

There is a torque on the wheel (mass of the weight x radius of the sprocket), and the equal and opposite reaction is the sprocket pulling on the chain. No opposite torque that I can see.

Can you have wheelie in this situation? Yes! The heavier the weight, the faster the bike accelerates. Eventually the center of mass of the bike plus rider is accelerated to the point where there is a force at the center of mass creating a moment about the rear wheel which will overcome the opposite moment of the mass holding the front wheel down. .

How much horizontal acceleration would that require for an actual car or bike? And assuming that the required horizontal acceleration can be achieved, how does the lifting torque from horizontal acceleration compare to the lifting torque from propulsion needed to achieve that acceleration, on an actual car or bike?

 

[sophiecentaur]

Is there something specific about the way that torque must be defined that makes it different than a moment created by a force?

I think this could be the basis of your problem and the answer is 'no'. If a force acts on an object then there will be a torque if the line of action of the force is not through the cm. That is the description of torque - force times perpendicular distance. The MI of the object will produce an equal and opposite torque, in the same way that a mass produces a reaction against a force, in a simple one dimensional situation. In a general two dimensional situation, an applied force will actually generate both a reaction force and a reaction torque so the bike will lift and move forward (we have seen them all do just that).
The MI of an object can be looked upon as the sum of all the effective contributions of all the parts of an object to the reactive torque it presents to an impressed torque. If the motor bike had just a single point mass, suspended on a light frame, above the ground, the two torques involved (action and reaction) might be a bit more obvious.