Question

For the normal force in the figure to have the same magnitude at all points on the vertical track, the stunt driver must adjust the speed to be different at different points. Suppose, for example, that the track has a radius of 2.9 m and that the driver goes past point 1 at the bottom with a speed of 16 m/s. What speed must she have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom?

Answer 1

According to the given conditions, her speed at the top of the vertical circle is

$$ \begin{aligned} v_{t} &=\sqrt{2 g r+v_{b}^{2}} \\ &=\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(2.9 \mathrm{~m})+(16 \mathrm{~m} / \mathrm{s})^{2}} \\ &=17.7 \mathrm{~m} / \mathrm{s} \end{aligned} $$