Worked exercise — Total bounce time (bilog)

Problem

Theory predicts that the total time $t_\text{tot}$ taken by a ball to stop bouncing, if dropped from a height $h_0$, follows a power law: $t_\text{tot} = a\cdot h_0^{\,1/2}, \qquad a = \frac{\sqrt{8k/g}}{1-k} \approx 6{,}76\ \text{s}/\sqrt{\text{m}} \quad (k = 0{,}875)$ The ball is dropped from five different heights and the total time is measured:

$h_0$ (m)	$t_\text{tot}$ (s)	$\log_{10} h_0$	$\log_{10} t_\text{tot}$
0{,}25	3{,}4	−0{,}60	0{,}53
0{,}50	4{,}6	−0{,}30	0{,}66
1{,}00	7{,}0	0{,}00	0{,}85
1{,}50	8{,}1	0{,}18	0{,}91
2{,}00	9{,}8	0{,}30	0{,}99

Verify on the bilogarithmic graph that the exponent is $1/2$ and work out $a$.

Solution

Set-up. The expected law $t_\text{tot} = a\,h_0^{1/2}$ is a power law: on the bilogarithmic graph (both axes on a log scale) it becomes a straight line whose slope is precisely the exponent. $\log_{10}(t_\text{tot}) = \log_{10}(a) + \tfrac{1}{2}\,\log_{10}(h_0)$

Slope = exponent. Taking the first and last point of the fourth and third columns, the slope is computed as: $n = \frac{\Delta\log_{10} t}{\Delta\log_{10} h_0} = \frac{0{,}99 - 0{,}53}{0{,}30 - (-0{,}60)} = \frac{0{,}46}{0{,}90} \approx 0{,}51$ consistent with the theoretical value $n = 1/2$.

Recovering $a$. The coefficient $a$ is read off at $h_0 = 1$ m, i.e. $\log_{10} h_0 = 0$: there the fit gives $\log_{10} t \approx 0{,}83$, hence $a = 10^{0{,}83} \approx 6{,}8\ \text{s}/\sqrt{\text{m}}$ in excellent agreement with the theoretical value $6{,}76\ \text{s}/\sqrt{\text{m}}$. $\ev{n \approx \tfrac{1}{2}, \qquad a \approx 6{,}8\ \text{s}/\sqrt{\text{m}}}$

Interpretation. The log-log slope coming out as $\approx 0{,}5$ confirms the dependence $t_\text{tot} \propto \sqrt{h_0}$: doubling the release height does not double the bounce time, it only increases it by a factor $\sqrt{2} \approx 1{,}41$.

On the log-log graph the five points line up; the slope $\approx 0{,}5$ is the exponent of the power law $t_\text{tot} = 6{,}76\,h_0^{1/2}$.

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Topics: Metodo sperimentale Skills: Uso delle scale logaritmiche · Linearizzazione dei dati · Lettura dei grafici