A $15-\mathrm{kg}$ ball of radius $4.0 \mathrm{~cm}$ is suspended from a point $2.94 \mathrm{~m}$ above the floor by an iron wire of unstretched length $2.85 \mathrm{~m}$. The diameter of the wire is $0.090 \mathrm{~cm}$, and its Young's modulus is $180 \mathrm{GPa}$. If the ball is set swinging so that its center passes through the lowest point at $5.0 \mathrm{~m} / \mathrm{s}$, by how much does the bottom of the ball clear the floor? Discuss any approximations that you make.
Call the tension in the wire $F_{T}$ when the ball is swinging through the lowest point. Since $F_{T}$ must supply the centripetal force as well as balance the weight,
$$
F_{T}=m g+\frac{m v^{2}}{r}=m\left(9.81+\frac{25}{r}\right)
$$
all in proper SI units. This is complicated, because $r$ is the distance from the pivot to the center of the ball when the wire is stretched, and so it is $r_{0}+\Delta r$, where $r_{0}$, the unstretched length of the pendulum, is
$$
r_{0}=2.85 \mathrm{~m}+0.040 \mathrm{~m}=2.89 \mathrm{~m}
$$
and where $\Delta r$ is as yet unknown. However, the unstretched distance from the pivot to the bottom of the ball is $2.85 \mathrm{~m}+0.080 \mathrm{~m}=2.93 \mathrm{~m}$, and so the maximum possible value for $\Delta r$ is
$$
2.94 \mathrm{~m}-2.93 \mathrm{~m}=0.01 \mathrm{~m}
$$
We will therefore incur no more than a $1 / 3$ percent error in $r$ by using $r=r_{0}=2.89 \mathrm{~m}$. This gives $F_{T}=277 \mathrm{~N}$. Under this tension, the wire stretches by
$$
\Delta L=\frac{F L_{0}}{A Y}=\frac{(277 \mathrm{~N})(2.85 \mathrm{~m})}{\pi\left(4.5 \times 10^{-4} \mathrm{~m}\right)^{2}\left(1.80 \times 10^{11} \mathrm{~Pa}\right)}=6.9 \times 10^{-3} \mathrm{~m}
$$
Hence, the ball misses by
$$
2.94 \mathrm{~m}-(2.85+0.0069+0.080) \mathrm{m}=0.0031 \mathrm{~m}=3.1 \mathrm{~mm}
$$
To check the approximation we have made, we could use $r=2.90 \mathrm{~m}$, its maximum possible value. Then $\Delta L=6.9 \mathrm{~mm}$, showing that the approximation has caused a negligible error.