Solution

The free-body diagrams of the student at the top and bottom of the Ferris wheel are
shown below. At the top (the highest point in the circular motion) the seat pushes up on
the student with a force of magnitude $$F_{N, top}$$, while the Earth pulls down with a force of
magnitude mg. Newton’s second law for the radial direction gives
$$mg-F_{N, top}=\dfrac{mv^2}{R}$$
At the bottom of the ride, $$F_N$$, the bottom is the magnitude of the upward force exerted by the
seat. The net force toward the center of the circle is (choosing upward as the positive
direction):
$$F_{N, top}-mg=\dfrac{mv^2}{R}$$
The Ferris wheel is “steadily rotating” so the value $$F_c=\dfrac{mv^2}{R}$$ is the same everywhere.
The apparent weight of the student is given by $$F_N$$.

(a) At the top, we are told that $$F_{N,top} = 556 N$$ and $$mg = 667 N$$. This means that the seat is
pushing up with a force that is smaller than the student’s weight, and we say the student
experiences a decrease in his “apparent weight” at the highest point. Thus, he feels
“light.”

(b) From (a), we find the centripetal force to be
$$F_{c}=\dfrac{mv^2}{R}=mg-F_{N,top}=667 N -556 N =111N$$
Thus, the normal force at the bottom is
$$F_{N,bottom}=\dfrac{mv^2}{R}+mg=F_c+mg=111 N+ 667 N =778 N.$$

(c) If the speed is doubled,$$F'_c=\dfrac{m(2v^2)}{R}=4(111 N)= 444 N$$ Therefore, at the highest
point we have
$$F'_{N,bottom}=F'_c+mg=444 N+667 N=1111 N= 1.11\times 10^3 N $$

Note: The apparent weight of the student is the greatest at the bottom and smallest at the
top of the ride. The speed $$v=\sqrt{gR}$$ would result in,$$ F_{N, top}=0$$ giving the student a
sudden sensation of “weightlessness” at the top of the ride.