Question

As depicted in Fig. 9-5, a ball $B$ is fastened to one end of a 24 -cm string, and the other end is held fixed at point $O$. The ball whirls in the horizontal circle shown. Find the speed of the ball in its circular path if the string makes an angle of $30^{\circ}$ to the vertical. The only forces acting on the ball are the ball's weight $m g$ and the tension $F_{r}$ in the cord. The tension must do two things: (1) balance the weight of the ball by means of its vertical component, $F_{T} \cos 30^{\circ} ;(2)$ supply the required centripetal force by means of its horizontal component, $F_{T} \sin 30^{\circ}$. Therefore, we can write $$ F_{T} \cos 30^{\circ}=m g \quad \text { and } \quad F_{T} \sin 30^{\circ}=\frac{m v^{2}}{r} $$ Solving for $F_{Y}$ in the first equation and substituting it in the second gives $$ \frac{m g \sin 30^{\circ}}{\cos 30^{\circ}}=\frac{m v^{2}}{r} \quad \text { or } \quad v=\sqrt{r g(0.577)} $$ However, $r=\overline{B C}=(0.24 \mathrm{~m}) \sin 30^{\circ}=0.12 \mathrm{~m}$ and $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$, from which $v=0.82 \mathrm{~m} / \mathrm{s}$.

          As depicted in Fig. 9-5, a ball $B$ is fastened to one end of a 24 -cm string, and the other end is held fixed at point $O$. The ball whirls in the horizontal circle shown. Find the speed of the ball in its circular path if the string makes an angle of $30^{\circ}$ to the vertical. The only forces acting on the ball are the ball's weight $m g$ and the tension $F_{r}$ in the cord. The tension must do two things: (1) balance the weight of the ball by means of its vertical component, $F_{T} \cos 30^{\circ} ;(2)$ supply the required centripetal force by means of its horizontal component, $F_{T} \sin 30^{\circ}$. Therefore, we can write $$
F_{T} \cos 30^{\circ}=m g \quad \text { and } \quad F_{T} \sin 30^{\circ}=\frac{m v^{2}}{r}
$$
Solving for $F_{Y}$ in the first equation and substituting it in the second gives
$$
\frac{m g \sin 30^{\circ}}{\cos 30^{\circ}}=\frac{m v^{2}}{r} \quad \text { or } \quad v=\sqrt{r g(0.577)}
$$
However, $r=\overline{B C}=(0.24 \mathrm{~m}) \sin 30^{\circ}=0.12 \mathrm{~m}$ and $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$, from which $v=0.82 \mathrm{~m} / \mathrm{s}$.

Added by Megan L.

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