A bead slides without friction around a loop-the-loop (Fig.). The bead is released from rest at a height h = 3.50R. How large is the normal force on the bead at point (A) if its mass is 50 g?

 

The speed at the top can be found from the conservation of energy for the bead-track-Earth system,
We define the bottom of the loop as the zero level for the gravitational potential energy.

$\mathrm{Since} {\mathrm{v}}_{\mathrm{i}} = 0, {\mathrm{E}}_{\mathrm{i}} = {\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{i}} = 0 + \mathrm{mgh} = \mathrm{mg}(3.50\mathrm{R})$

The. total energy of the bead at point (A) can be written as ${\mathrm{E}}_{\mathrm{A}} = {\mathrm{K}}_{\mathrm{A}}+{\mathrm{U}}_{\mathrm{A}} = \frac{1}{2}{{\mathrm{mv}}_{\mathrm{A}}}^2+\mathrm{mg}\left(2\mathrm{R}\right)$

Since mechanical energy is conserved, ${\mathrm{E}}_{\mathrm{i}} = {\mathrm{E}}_{\mathrm{A}}, \mathrm{we} \mathrm{get}$

$\mathrm{mg}(3.50\mathrm{R}) = \frac{1}{2}{{\mathrm{mv}}_{\mathrm{A}}}^2+\mathrm{mg}\left(2\mathrm{R}\right)$

simpliflying, ${{\mathrm{v}}_{\mathrm{A}}}^2 = 3.0 \mathrm{gR} \Rightarrow {\mathrm{v}}_{\mathrm{A}} = \sqrt{3.0 \mathrm{gR}}$

To find the normal force at the top, we construct a force diagram as shown,

Here we assume that N is downward, like mg.

$\sum _{ }{\mathrm{F}}_{\mathrm{y}} = {\mathrm{ma}}_{\mathrm{y}}; \mathrm{N} + \mathrm{mg} = \frac{{\mathrm{mv}}^2}{\mathrm{r}}$

$\mathrm{N} = \mathrm{m}[\frac{{\mathrm{v}}^2}{\mathrm{R}}-\mathrm{g}] = \mathrm{m}[\frac{3.0\mathrm{gR}}{\mathrm{R}}-\mathrm{g}] = 2.0 \mathrm{mg}$

$\mathrm{N} = 2.0(50\times {10}^{-3}\mathrm{kg})(10.0 \mathrm{m}/{\mathrm{s}}^2)$

   = 1.0 N downward