Newtons Laws EX 23

The diagram shows a segment of her path and herself on the seat. The free-body diagram shows the forces acting on the woman. There is the force of gravity that acts down and the normal force that acts up. Her apparent weight is equal to the normal force.

(a) She feels weightless at the top when the normal force is zero which is when the force of gravity alone acts to accelerate her downward so that she moves in a circle of radius 20 m. The acceleration points towards the center of the path. In the diagram the normal force is shown as it would be there in a general case. The fre body diagram is shown with a positive y-axis pointing up. Consequently, the y-component of her acceleration is negative and it is equal to $-v^{2} \over r$. In general, we write

${\overrightarrow {F_{N}}}+{\overrightarrow {F_{G}}}=m{\overrightarrow {a}}$

If $F_{N}=0$, then

$-mg=m({-v^{2} \over r})$

$gr=v^{2}$

${\sqrt {gr}}=v$

$v={\sqrt {10\times 20}}$

$v=14$ m/s

Note that the value of $m{v^{2} \over r}$ is equal to her weight, 600 N.

(b) At the bottom, the normal force must provide enough force to overcome the gravitational force and accelerate the woman upward toward the center of the Ferris wheel. In this case, the y-component of her acceleration is positive and it is equal to $v^{2} \over r$.

${\overrightarrow {F_{N}}}+{\overrightarrow {F_{G}}}=m{\overrightarrow {a}}$

$F_{N}-mg=m({v^{2} \over r})$

$F_{N}-600=600$

$F_{N}=1200$ N