Angular and linear displacement of a turning car My illustration is supposed to show a simplified car (with just two wheels, where the front wheel has been rotated $-90 \text{ degrees}$). Let the car travel at constant speed. This means that the only horizontal force on the car is the centripetal force, which acts towards the rear wheel since we're rotating around it. The result of this is that point $P$ in the middle of the car will move along the half circle I've drawn.

I'm wondering how this is possible, since we just have radial acceleration (which should only change the direction), but obviously, as the car moves, both its angular and linear displacement are going to change.

Based on the centripital force $F_c = \frac{mv^2}{r}$, where $r$ is the distance from the wheels, how is the linear displacement of the outer wheel and the angular displacement about $P$ calculated?

Am I thinking about this the wrong way? A part of me also thinks torque should be involved.

• Acceleration is always perpendicular to velocity here, so the speed of the outer wheel remains constant. This is because there exists no component of acceleration in the direction tangent to the path of the outer wheel; all the acceleration goes into changing the direction of the velocity. Because tangential speed will remain constant, angular speed will remain constant. It follows that angular and linear displacement will increase linearly. Note that if a torque were applied in the direction of tangential velocity, the tangential speed of the outer wheel would change over time. May 4 '15 at 20:22
• Regarding calculations, remember that $\theta = \frac{s}{r}$, which leads to $\omega = \frac{v}{r}$ and $\alpha = \frac{a}{r}$. May 4 '15 at 20:33