enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ – Rations May 4 '15 at 20:22
  • $\begingroup$ Regarding calculations, remember that $\theta = \frac{s}{r}$, which leads to $\omega = \frac{v}{r}$ and $\alpha = \frac{a}{r}$. $\endgroup$ – Rations May 4 '15 at 20:33

As the car turns around, since the car is a rigid body(ideally), not only will the rear wheel experience a centripetal force, but also other parts of the car. You can illustrate this by recalling your experience in a car when the car turns around. You will feel a force pushing you in the direction of rotation. So every part of the car is experiencing a centripetal force and doing a circular motion. Also, I think you do not mean to say angular displacement and linear displacement, object in circular motion has both angular and linear displacement. Do you mean motion in centripetal and tangent directions?

  • $\begingroup$ The thing is, I'm programming a simulation. And P (the car) is represented by a positional vector (xy-plane), and a rotation. Initally the car is facing to the right with 0 degree rotation. So I need to be able to calculate its new rotation (angular displacement?), and its new position given the centripetal acceleration and its velocity (and angular velocity?). $\endgroup$ – lawls May 5 '15 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.