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Okay, so I got some really great advice from the community here but I am still hung up on some conceptual problems. I posted earlier trying to determine the axis that a bicycle rotates about when doing a wheelie. Since then I have realized that there is not any unique axis in rotational motion, rather axes are better chosen for the convenience of calculation. Here is the diagram of the object of interest:

enter image description here

It shows the frame of a bike and its front tire at the moment that it begins to lift off the ground. As I have chosen the bike as a frame of reference, its acceleration creates fictitious inertial forces in both the x and y direction.

I have chosen the center of mass as having the same position as the axis of rotation in order to easily break down motion into its transnational and rotational components.

I can calculate the moment that the body begins to rotate about its center of mass by summing the torques produced by the normal force on the rear axle and the horizontal applied force on the back of the frame. When the applied force produces a large enough torque there will be an angular acceleration about the center of mass.

What I fail to understand is why the center of mass rises above its initial height. What accelerates the COM to move it above its initial height with respect to the surface it is moving across?

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  • $\begingroup$ Fa is produced by the torque at the back wheel, but is simply a force acting in the forward direction $\endgroup$ – Lance Bowley May 3 '15 at 18:12
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So the torque from the back wheel causes a net torque about the centre of mass counterclockwise. This means the back wheel exerts a greater reaction force on the ground so when we resolve forces the reaction force exceeds mg, causing a net force up on the centre of mass.

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  • $\begingroup$ How could you calculate how much net force is being applied upward by the back wheel? $\endgroup$ – Lance Bowley May 3 '15 at 18:45
  • $\begingroup$ In these sorts of problems the only things which matter are working out net torque and force on centre of mass. F=ma and T=I(alpha). Wheel produces frictional force from ground and reaction force and CoM produces gravitational force. The torque produced by wheel is a given. Linear Acceleration of back wheel is zero so if we choose this to take moments about we see (Torque times wheel radius)-(distance to Com*Mg)=I(alpha)=(distance to Com)^2*M(alpha).You could also take moments about the Com and solve simultaneous equations for F=ma and T=I(alpha) $\endgroup$ – user63826 May 3 '15 at 19:08
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Sorry if I'm missing a detail here - but how can you choose the center of mass? Isn't that determined by the shape/mass distribution of the bike? And the axis of rotation ought to be the center of the rear tire, no?

Just imagining a wheelie I can see that the center of mass should move upwards to some degree, as the wheelie is an upward rotation.

(posting as answer because I don't have the rep to comment...)

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  • $\begingroup$ Yeah that is pretty silly that you cannot comment yet, if you ask me. When I said I chose the center of mass, what I meant to say is that I chose the axis of rotation to be at the same location as the COM $\endgroup$ – Lance Bowley May 3 '15 at 18:13

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