As drawn in Fig. 9-6, a 20-g bead slides from rest at $A$ along a frictionless wire. If $h$ is $25 \mathrm{~cm}$ and $R$ is $5.0 \mathrm{~cm}$, how large a force must the wire exert on the bead when it is at $(a)$ point- $B$ and $(b)$ point- $D ?$
(a) As a general rule, remember to keep a few more numerical figures in the intermediate steps of the calculation than are to be found in the answer. This will avoid round-off errors. Let us first find the speed of the bead at point- $B$. It has fallen through a distance $h-2 R$ and so its loss in $\mathrm{PE}_{\mathrm{G}}$ is $m g(h-2 R)$. This must equal its $\mathrm{KE}$ at point- $B$ :
$$\frac{1}{2} m v^{2}=m g(h-2 R)$$
where $u$ is the speed of the bead at point- $B$. Hence,
$$v=\sqrt{2 g(h-2 R)}=\sqrt{2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.15 \mathrm{~m})}=1.716 \mathrm{~m} / \mathrm{s}$$
As shown in Fig. $9-6(b)$, two forces act on the bead when it is at $B$ : (1) the weight of the bead $m g$ and (2) the (assumed downward) force $F$ of the wire on the bead. Together, these two forces must supply the required centripetal force, $m u^{2} / R$, if the bead is to follow the circular path. Therefore, write
$$m g+F=\frac{m v^{2}}{R}$$
or
$$F=\frac{m v^{2}}{R}-m g=(0.020 \mathrm{~kg})\left[\left(\frac{1.716^{2}}{0.050}-9.81\right) \mathrm{m} / \mathrm{s}^{2}\right]=0.98 \mathrm{~N}$$
The wire must exert a $0.98$ N downward force on the bead to hold it in a circular path.
(b) The situation is similar at point- $D$, but now the weight is perpendicular to the direction of the required centripetal force. Therefore, the wire alone must furnish it. Proceeding as before,
$$v=\sqrt{2 g(h-R)}=\sqrt{2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.20 \mathrm{~m})}=1.98 \mathrm{~m} / \mathrm{s}$$
or
$$
F=\frac{m v^{2}}{R}=\frac{(0.020 \mathrm{~kg})\left(1.98 \mathrm{~m}^{2} / \mathrm{s}^{2}\right)}{0.050 \mathrm{~m}}=1.6 \mathrm{~N}
$$