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When the person is at the top of the ferris wheel $mg$ would be greater than $F_{Normal}$. Hence the person would be accelerating downwards. So why does the person move towards the center of the circle as if he was in freefall? Since Newtons second law is "If there is a force acting on the system, the system would move in the direction of the net force in a manner that $F_{net}=ma$" Similarly when he is at the bottom why does he not accelerate upwards since the normal force is > than mg.


3 Answers 3


If there is a force acting on the system, the system would move in the direction of the net force in a manner that $F_{net}=ma$

I think you are confusing motion and acceleration. Forces determine the acceleration of an object, not the motion at that instant.

The car at the top is accelerating downward, and the car at the bottom is accelerating upward. But they also have velocity in the sideways direction


Actually what happens in situations like these is well explained by the term apparent weight. The direction of the force depends on acceleration which in turn depends on the change in velocity. The key point is that the human at the top is not in equilibrium but is in accelerated motion.

Notice that there only two forces acting on the human at the top. There is the gravitational force pulling down and the seat (the normal force) pushing up. These two forces do not have equal magnitude because the human is not in equilibrium but instead accelerating by moving in a circle. The direction of the acceleration at this instant is downward toward the center of the circle.

Using Newton's law :

N - mg = ma(a in y direction) = m(-v^2)/R

N = m(g - v^2/R)

This says that at the top, your apparent weight will be lower than your actual weight by some amount that depends on both the size of the wheel and the angular velocity.


The body is going round in a circle and hence undergoing centripetal acceleration produced by the net force due to the two forces acting on the body of mass $m$, the weight of the body $mg$ and the normal reaction of the seat on the body $F_{\rm normal}$.

At the top assuming down is positive and using Newton's second law $mg -F_{\rm normal, top} =\dfrac{mv_{\rm top}^2}{R}$ where $v_{\rm top}$ is the speed of body at the top of the Ferris wheel and $R$ the radius of the Ferris wheel.
Notice that if the Ferris wheel roared fast enough the normal force could be zero or even acting downwards, the normal force in the case being provided by the seat belt.

At the bottom with up as positive, $F_{\rm normal, bottom}-mg =\dfrac{mv_{\rm bottom}^2}{R}$


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