Motorcycle Wheelie (Angular Momentum)

[Osh]

Homework Statement

The distance between the centers of the wheels of a motorcycle is 155 cm. The center of mass of the motorcycle, including the biker, is 88.0 cm above the ground and halfway between the wheels. Assume the mass of each wheel is small compared with the body of the motorcycle. The engine drives the rear wheel only. What horizontal acceleration of the motorcycle will make the front wheel rise off the ground?

Homework Equations

[tex]\tau=rFsin\phi[/tex]
[tex]\sum\tau=I\alpha[/tex]


I am having problems visualizing this question. Any way I look at it, I can't seem to figure out how a horizontal force on the back wheel of the motorcycle would cause the front wheel to move vertically.

If I look at it as the rotation of a rigid object, the pivot point would have to be the centre of the back wheel. The torque from the gravity at the centre of mass would have to somehow be negated, right? But the only other force acts horizontally, meaning there would be no force in the vertical direction.

 

[LowlyPion]

Welcome to PF.

I think you are almost grasping it. The Torque of the motor on the backwheel is what will lift the bike. But the torque of the motor also generates the F = m*a of the overall system (bike/rider).

If you generate forward acceleration in terms of g great enough, then you will overcome the weight acting through the center of mass about the back wheel hub won't you?

 

[Osh]

I don't see how it would do that, though. I envision something like this:

Where F_g is the force of gravity and F_a is the force from the acceleration.

So in order for the wheel to lift, there would have to be a torque acting counterclockwise from the pivot of the back wheel, right? But I don't see a force with a component in that direction.

 

[LowlyPion]

Look at where the center of mass is. It is .88 m high and .775 m from where it acts to ground. Ultimately the pivot point is the tire and ground isn't it? The back wheel is rotating clockwise for forward motion. But for every action there is a reaction.

So doesn't it seem that the center of mass acting about the .88m height backward, taking into account that the bike is accelerating of course, needs to balance the downward weight acting through the .775m to ground?

 

[djeitnstine]

I don't see how it would do that, though. I envision something like this:

Where F_g is the force of gravity and F_a is the force from the acceleration.

So in order for the wheel to lift, there would have to be a torque acting counterclockwise from the pivot of the back wheel, right? But I don't see a force with a component in that direction.

Remember [tex][F_{a}[/tex] is converted to torque via the chain (Also remember Newton's 3rd Law)

 

[Osh]

Oh, I think I see it now: because the back wheel is interacting with the body of the bike through the chain, an equal and opposite force to the force of acceleration on the back wheel is applied to the centre of mass. Essentially, F_a is in the opposite direction as in the picture. So, I can set the bottom of the back wheel to be the pivot point and calculate the torque about that point with something like this:

[itex]\sum\tau=0[/itex]
[itex]F_{g}r\sin\theta=F_{a}r\sin\phi[/itex] where [tex]\theta[/tex] is the angle between F_g and the radius, and [tex]\phi[/tex] is the angle between F_a and the radius
[tex]a=g\sin\phi/\sin\theta[/tex]

Am I on the right track?

 

[LowlyPion]

Since what they do give is the effective distance from the point on the ground that the CoM is acting down on, as well as the CoM and the height above the wheel that the bike has to put into the torque that will accelerate the bike, you already have the projection of the moment arms where the forces are perpendicular.

To balance the moments then just before front wheel lifts then doesn't it simply look like .88*m*a = .775*m*g

 

[djeitnstine]

Oh, I think I see it now: because the back wheel is interacting with the body of the bike through the chain, an equal and opposite force to the force of acceleration on the back wheel is applied to the centre of mass. Essentially, F_a is in the opposite direction as in the picture. So, I can set the bottom of the back wheel to be the pivot point and calculate the torque about that point with something like this:

[itex]\sum\tau=0[/itex]
[itex]F_{g}r\sin\theta=F_{a}r\sin\phi[/itex] where [tex]\theta[/tex] is the angle between F_g and the radius, and [tex]\phi[/tex] is the angle between F_a and the radius
[tex]a=g\sin\phi/\sin\theta[/tex]

Am I on the right track?

Correct

 

[Osh]

Apparently my professor used a method in which he used the reaction force from the ground on the back wheel to create a torque around the centre of the back wheel, causing the bike to rotate at a certain acceleration. I think I'll just start over that way. Thanks for your help, though!

 

[LowlyPion]

Apparently my professor used a method in which he used the reaction force from the ground on the back wheel to create a torque around the centre of the back wheel, causing the bike to rotate at a certain acceleration. I think I'll just start over that way. Thanks for your help, though!

My post #7 assumes that approach.

 

[djeitnstine]

Apparently my professor used a method in which he used the reaction force from the ground on the back wheel to create a torque around the centre of the back wheel, causing the bike to rotate at a certain acceleration. I think I'll just start over that way. Thanks for your help, though!

You were doing the same thing. I don't see why you have to start again :p

Pion was also correct. Everyone was correct!

 

[Osh]

You were doing the same thing. I don't see why you have to start again :p

Pion was also correct. Everyone was correct!

Well, that's confusing. In the professor's approach, there was no reaction force on the centre of mass acting backwards. I don't see how they would yield the same answer, considering that.

Let me see how it works:

For the torque from the reaction force on the bottom of the back wheel...

[tex] \tau_{R} = F_{a}R [/tex] where F_a is that force and R is the radius of the wheel
[tex] \tau_{R} = MaR [/tex] where M is the mass being accelerated​

So, M is the mass of the bike, right? Anyway, the torque of the weight at the centre of mass can be taken into account:

[tex] \tau_{CM} = Mgl/2 [/tex] with l being the length between wheels

[tex]\sum\tau = 0[/tex] so the torques have to be equal:
[tex]MaR = Mgl/2[/tex] The masses cancel:
[tex]a = gl/2R[/tex]​

Hm. The radius of the wheel isn't given. I don't understand why you used the height, Pion. I must have done some wrong. Assuming that you're right, though, then both methods would yield the same answer somehow.

I'll take your word for it, djeitnstine, that everyone's correct. I don't want this question to cause me anymore grief. Mind you, it did get me thinking about the material, which is what homework's suppose to do. (: