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Say for example you have a person riding a Ferris wheel. What is the difference in his acceleration (towards the center) for when he is at the highest point to when he is at the lowest point? So the answer is that his acceleration is the same every where, that is it is the centripetal acceleration.

My question is, why isn't his net acceleration at the top g+Ca, towards the ground, and his acceleration at the bottom g-Ca, towards the sky?

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The centripetal force is not a particular force; it is whatever force causes the circular motion. The only forces acting on the rider are weight (i.e. $mg$) and contact forces (normal and friction) from the car/seat.

Because the motion is uniform and circular you know that these add up to the value of the centripetal force. Because you know the weight at all times you can compute the normal forces if you want.

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Actually that is exactly what is happening. The net acceleration is radial toward the center at all points on the circular travel. His acceleration magnitude is the same at all points but the direction of the acceleration vector is constantly changing.

If you expressed the forces in vector terms, it would always be $\vec{a}$ = $\vec{g}$ + $\vec{C}$ but because the directions are opposite for the direction of $\vec{C}$ at top and bottom, the acceleration at the bottom could be expressed as $\parallel{g + C}\parallel$ and at the top as $\parallel{g - C}\parallel$ if you are using scalar magnitudes. The $\vec{C}$ component is "rotating" as time progresses. At points in between the top and bottom points, there would be a dot-product trig term to do the vector addition.

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Supposing that the ferris wheel turns at a constant speed, your acceleration everywhere along the ride has to be $\frac{v^2}{r}$ directed towards the center (otherwise you would fly off). However, the forces acting on you will change as you go around the circle.

The forces are:

The gravitational force downward: $-m g \hat{\mathbf{y}}$

The normal force of your seat acting upwards on your bottom: $N_y \hat{\mathbf{y}}$

The frictional force of your seat acting forwards/backwards on your bottom: $f \hat{\mathbf{x}}$

The normal force of your backrest acting on your back: $N_x \hat{\mathbf{x}}$

Then $$\sum F_x = m a_x = - m\frac{v^2}{r} \cos\theta$$

$$\sum F_y = m a_y = -m\frac{v^2}{r} \sin\theta$$

where of course $\theta$ is your angle wrt the ground. From this you can figure out what some of the other forces need to be to keep you in your seat. For example, $N_y$ is least at the top and greatest at the bottom.

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