# HW 9 pROBLEM 8

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The distance between the centers of the wheels of a motorcycle is 173 cm. The center of mass of the motorcycle, including the rider, is 97.5 cm above the ground and halfway between the wheels. Assume the mass of each wheel is small compared to 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?

Draw a careful diagram to illustrate what is happening when the front wheel just lifted off the ground (touching but no forces). Label all the forces and useful parameters. Carefully observe where the different forces are applied so you can figure out the torques. Identify the center of mass clearly and set it as the rotational axis. The vector R is a vector starting from the center of mass to the point on the rear wheel where the frictional and normal forces are applied.

Here the example is simplified with the physics perspective and may seems unsatisfying regarding details of component level interaction forces and torques from an engineering view point. But let's treat it as a simplified physics case.

When the front wheel just lifted from the ground, there are three forces applied on the bike -- see the diagram. Taking the center of mass as the rotational axis, the gravitational force won't create torque, while the frictional and normal forces will. Define CCW as the positive direction, and set the total torque to be zero -- the bike isn't spinning up or down with an angular acceleration:

$$\tau _{net}=\tau _f -\tau _N=0$$

$$f \cdot 0.975-N\cdot 0.865=0$$

$$\therefore f=0.887mg.$$

Here we used $N=mg$ since the bike is not accelerating in the vertical direction. We also used $\tau=R_\bot F$ to calculate the torque.

Then considering the net force in the horizontal direction, there is only $f$, which gives $f=ma$. Thus you can find $a$ based on the relation derived from the torque equation.

answered Oct 20, 2013 by (21,750 points)
edited Oct 20, 2013