Past exam of the physics A course of the University of Cambridge / 2022 / Paper 1 / Section C

The original apparatus has two small masses at distance $L/2$ from the suspension axis. Neglecting the rod's moment of inertia, their total moment of inertia is $I=2m(L/2{)}^2=m{L}^2/2$, consistent with the supplied torsion pendulum period. Treat $m$ and $L$ as exact because no uncertainties are specified. Rearranging the period relation givesThe torsion constant is a restoring torque per unit angle. In dimensional calculations the radian is dimensionless.

Using first-order propagation of uncertainty, with the quoted period uncertainty interpreted as a standard uncertainty,Consequently

Rearrange the given torsion balance equilibrium relation asA useful choice, preserving equal measurement uncertainties on the vertical axis, isWith angle measured in radians, $y$ is dimensionless and $x$ has units ${\mathrm{m}}^{-2}$, so $c$ has units ${\mathrm{m}}^2$, as required. The four $x$ values, in the order of the original table, areThe corresponding $y$ values are $(0.066,0.055,0.044,0.025)$, each with standard uncertainty ${\sigma }_y=0.002$. The uncertainty of $x$ is negligible by assumption, making linear regression through the origin appropriate for the proposed model.

All four angle measurements have the same standard uncertainty, so ordinary least squares minimizesDifferentiation givesThus the unique minimum isThe overbars denote sample means over the four observations. Numerically,soThe fixed $x_i$ have negligible uncertainty. Assuming independent errors in the four $y_i$, propagation of uncertainty in quadrature givesHenceIf the period measurement is independent of the angle measurements, including its calibration uncertainty givesThis last uncertainty includes the stated random measurement errors only. Whether the model describes the data must be checked before treating it as a reliable measurement of the gravitational constant.

Assume independent normally distributed angle errors with known standard deviation ${\sigma }_y=0.002$. The relevant chi-squared test uses the sum of squared standardized residuals,For the model through the origin, the predictions and residuals, expressed in milliradians, areConsequently ${\chi }_0^2=30.0280$. There are $4-1=3$ statistical degrees of freedom. The upper-tail P-value is approximately $1.36\times 10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.

For the alternative linear regression, minimize ${\sum }_i(y_i-c^{\prime }{x}_i-d{)}^2$. Setting both partial derivatives to zero gives the normal equations, whose solution isFor these data $\bar{x}=2508.83699{\mathrm{m}}^{-2}$, $\bar{y}=0.0475$, and $S_{xx}=4.830232\times 10^6{\mathrm{m}}^{-4}$. This yieldsThe new predictions are $(66.6548,55.4171,41.4315,26.4966)\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\mathrm{mrad}$. ThereforeThe affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta {\chi }^2=27.668$ has a chi-squared distribution with one statistical degree of freedom under the nested zero-intercept hypothesis, corresponding to $p\simeq 1.44\times 10^{-7}$. Thus the model with a fitted intercept provides the better description, rather than merely lowering the residual by adding a parameter.

With the measurement variance treated as known, the coefficient uncertainties areHence $d=(13.11\pm 2.49)\mathrm{mrad}$, about $5.26$ standard deviations above zero. A positive constant systematic error in the angular zero, or an additional nearly constant torque, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred gravitational constant and cannot remove this residual pattern.

Using the fitted slope after allowing for this offset would give $G^{\prime }=\kappa {\hat{c}}^{\prime }/(mML)=(7.27\pm 0.51)\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.