Bifilar suspension lab report

1. Title and aim

Moment of inertia by bifilar suspension. Determine the moment of inertia of a uniform slender rod about its vertical centre-of-mass axis from bifilar oscillations at different string lengths.

2. Abstract

Five observations were generated from the ideal teaching model over the settings listed below. A straight-line fit gave an estimate of moment of inertia of 0.01042 kg m², differing by 0.026% from the model value. The reported scatter comes from scale rounding and any stated model approximation, not measured laboratory noise.

3. Apparatus and fixed settings

• Uniform slender rod: Supply the rotating test body: mass M = 0.500 kg and rod length D = 0.500 m.
• Two strings and rigid overhead support: Suspend the rod horizontally using equal strings and matching top and bottom separation 2a = 0.200 m.
• Rule and balance: Measure equilibrium string length L, half-separation a, rod length D, and rod mass M.
• Timer and angular marker: Time 20 small torsional cycles using a consistent orientation and direction.
• Uniform slender rod: mass M = 0.500 kg; rod length ℓ = 0.500 m.
• Half-separation a = 0.100 m; full separation d = 2a = 0.200 m.
• Reference gravity g = 9.80665 m s⁻²; example yaw amplitude 3°.

4. Theory and graph choice

Suspend a horizontal rod from two equal parallel strings and twist it gently about the vertical axis through its centre. Use changes in period with string length to infer rotational inertia. The virtual rod has mass 0.500 kg, length 0.500 m, and suspension points 0.200 m apart. The animation is accelerated and the timing observations are generated, not physical measurements.

The calculation fits y = mx + b in SI units. The graph may display scaled axes for readability; those display factors are removed before calculating the physical result. The gradient has units s² m⁻¹.

5. Method

1. Record M = 0.500 kg, D = 0.500 m and theoretical I = MD squared/12. Identify the vertical axis through the centre.
2. Set equal string lengths L. Keep the rod horizontal, with upper and lower attachment separations both 0.200 m and centred about the rod midpoint.
3. Twist the rod only a few degrees about its central vertical axis and release without a push. Avoid lateral swinging and rocking.
4. Start at a chosen orientation moving in one direction and stop after 20 returns in the same direction. Record the generated time; repeat if available.
5. Use at least five well-spaced values of L while retaining the same rod and attachment spacing. Recheck equal lengths and horizontality each time; calculate T = mean time/20 and T squared.
6. Fit T squared = mL + b. Calculate I = mMga squared/(4pi squared), inspect intercept and residuals, and compare with 0.0104167 kg m squared with suitable uncertainty discussion.

6. Observations and calculation table

All observations use fixed model settings. Values are rounded only at the instrument reading stage; further calculations use full precision. Repeated physical trials are not claimed.

Download the sample observations (CSV) ↓

7. Worked calculation for one observation

8. Graph, gradient and final calculation

Dots show the sample observations. The line is the ordinary least-squares fit.

9. Uncertainty and comparison

Percentage difference = |estimate − model value| / |model value| × 100 = 0.0257%. Percentage difference is not experimental uncertainty.

For n = 5 observations, Σresidual² = 7.0857e-07; s² = Σresidual²/(n − 2). Gradient standard error = √[s²/Σ(x − x̄)²] = 0.00102. Intercept standard error = √[s²(1/n + x̄²/Σ(x − x̄)²)] = 6.5203e-04.

Propagating only the relevant fit coefficient gives a fit-only standard error of 1.2725e-06 kg m². This is a one-standard-error quantity under the regression assumptions. Deterministic rounded model data do not supply a complete physical uncertainty budget.

The inferred inertia depends on mass, string length, timing and the square of half-separation a. Independent fractional uncertainties approximately satisfy (uI/I)² = (uM/M)² + (uL/L)² + (2uT/T)² + (2ua/a)² + (ug/g)² for a single reading. Fit uncertainty alone excludes these geometrical and calibration terms.

10. Result and discussion

The observations follow the stated model relationship. Check whether the intercept and residual pattern suggest a systematic offset or a missing approximation. Agreement with the model does not independently verify its input constants.

11. Precautions and limitations

• Keep equal string lengths and matching symmetric upper and lower attachment separations.
• Use half-separation a = 0.100 m, not the full 0.200 m separation, in the stated formula.
• Use small twists and reject runs showing lateral swing, rocking, slipping attachments or slack strings.
• Keep clamps secure and distinguish accelerated generated timing from measurements on a real apparatus.

The model assumes a rigid uniform slender rod, equal massless inextensible strings, symmetric attachment points, matching upper and lower separation, a horizontal equilibrium rod and small isolated torsional motion. It ignores string torsional stiffness, damping, attachment inertia, finite rod width, mode coupling and support movement. The reference g is 9.80665 m/s squared. Generated readings and accelerated animation cannot reproduce the alignment difficulties or measurement uncertainty of a physical suspension. A finite-width bar requires the appropriate inertia formula; the half-separation convention must be checked before comparing with any other apparatus. Equal, parallel strings are attached symmetrically about the centre of mass. The model uses the small-yaw relation T = 2π√[IL/(Mga²)]. It assumes negligible string mass, rod width, damping and support compliance. The simulated target inertia is Mℓ²/12. Angles above 5° cannot be recorded.

12. Conclusion

The worked example illustrates how to obtain moment of inertia from measurements, graph variables and a fitted coefficient. The estimate is 0.01042 kg m². A real experiment must additionally evaluate the apparatus-specific uncertainties and corrections described above.