3. Theory
3.1 Rotating unbalance
A mass m whose centre of mass lies at radius r from the axis of a shaft turning at ω rad/s travels in a circle. The shaft supplies the centripetal force, and the mass pulls on the shaft with an equal and opposite force that turns with it:
\[ F = m\,r\,\omega^2 \]
For balancing, a block is described completely by its moment mr and its angle. On this bench mr is measured in units: one unit is the moment of 1 g hanging on the effective radius R of the pulley (section 3.3), so 1 unit = R g·mm, about 25.5 g·mm. Balancing works with ratios, so units serve for every calculation except those that need absolute forces (Experiments 5 and 7).
3.2 Static balance and the triangle of moments
A shaft is statically balanced when it stays where it is put at any angle. The weight of a block with moment mr at angle θ exerts a torque about the shaft that depends on mr and θ, and the torques of all the blocks cancel at every angle only if their moment vectors add to zero:
\[ \sum m_i r_i \cos\theta_i = 0, \qquad \sum m_i r_i \sin\theta_i = 0 \]
Drawn head to tail, the vectors form a closed polygon. Three blocks of known moment form a triangle with fixed sides, so their relative angles are fixed. With block a at 0°, the other two must close the vector \(\mathbf{R} = -m_a r_a\angle 0°\). The cosine rule gives the angle α between block b and \(\mathbf{R}\):
\[ \cos\alpha = \frac{(m_b r_b)^2 + R^2 - (m_c r_c)^2}{2\,m_b r_b\,R}, \qquad \theta_b = \theta_R \pm \alpha \]
and block c closes the triangle. The ± gives two solutions, the triangle drawn on one side of \(\mathbf{R}\) or the other. Both balance the shaft statically, and the positions of the blocks along the shaft play no part.
3.3 Measuring a moment with the pulley
With the belt off, one block on the shaft and a hanger of mass m on a cord round the extension pulley, the shaft turns until the weight of the block balances the hanger. If φ is the angle the block has turned up from hanging straight down,
\[ m\,R = m_b r_b \sin\varphi \pm \frac{T_f}{g} \]
R is the effective radius of the pulley, the groove radius plus half the cord diameter, \(R = D/2 + d/2\), and \(T_f\) is the friction torque of the bearings. While you add mass, friction holds the shaft back and the + sign applies. While you take mass off, friction holds it up and the − sign applies. With mr in units (grams at R), a plot of m against sin φ is a straight line of slope mr:
\[ m = (mr)\sin\varphi \pm m_f, \qquad m_f = \frac{T_f}{g\,R} \]
The loading line lies \(2m_f\) above the unloading line, and the mean of the two removes friction. Tapping the mountings lets the shaft creep towards the friction-free line. The block cannot be held beyond 90°: once m exceeds \(mr + m_f\), the load runs the shaft over the top. φ is measured from the block's unloaded protractor reading \(\varphi_0\), not from the zero of the scale.
3.4 The release test
To test static balance, turn the shaft to an angle and let go. A net unbalance U whose heavy side lies at angle φ from the bottom exerts a torque \(U g \sin\varphi\). The shaft moves only if this exceeds the breakaway friction torque \(T_s\) of the bearings, a little larger than the running friction. A shaft whose unbalance is smaller than \(T_s/g\) passes the test from every angle, and a small unbalance shows only when the shaft is released with its heavy side near the horizontal. Release from several angles round the full circle. The belt adds the drag of the motor to the friction and hides almost any unbalance, so it must come off for this test.
3.5 Dynamic balance
When the shaft runs, each block exerts a rotating force \(m r \omega^2\) in its own plane. The shaft runs without vibration at every speed only if
- there is no resultant force: \(\sum m_i r_i \angle\theta_i = 0\), the condition for static balance, and
- there is no resultant couple: \(\sum m_i r_i\,l_i \angle\theta_i = 0\), where \(l_i\) is the axial position of block i measured from a reference plane.
Taking the reference plane through mass 1 removes mass 1 from the couple equation. A dynamically balanced shaft is also statically balanced, but the reverse does not hold. Two equal masses at 180° and a distance s apart satisfy condition 1 but not condition 2: their forces form a couple \(m r\,s\,\omega^2\) that rocks the shaft, so the two bearings move in opposite directions.
Two masses balance dynamically only when they lie in the same plane. Three masses spread along the shaft balance only when the middle one is at 180° to the outer two, with \(F_2 = F_1 + F_3\) and \(a_2 F_2 = a_3 F_3\) about mass 1. Two adjacent blocks act as a single mass at their common plane.
3.6 The four-mass problem, by drawing and by calculation
Masses 1 and 2 are known in moment, angle and position. Masses 3 and 4 are known in moment only. Four unknowns remain: \(\theta_3, \theta_4, l_3, l_4\).
Angles from the mr polygon. Draw \(m_1r_1\) vertically, add \(m_2r_2\) at its angle, and call the vector that closes the polygon back to the start \(\mathbf{R} = -(m_1r_1\angle\theta_1 + m_2r_2\angle\theta_2)\). Swing an arc of radius \(m_3r_3\) from the start of \(\mathbf{R}\) and an arc of radius \(m_4r_4\) from its end. The arcs cross at two points, and each gives a pair of angles. By calculation, \(\theta_3 = \theta_R \pm \alpha\) with α from the cosine rule of section 3.2, and then \(m_4r_4\angle\theta_4 = \mathbf{R} - m_3r_3\angle\theta_3\).
Positions from the mrl polygon. With the reference plane through mass 1, the couple condition reads \(m_3r_3l_3\angle\theta_3 + m_4r_4l_4\angle\theta_4 = -m_2r_2l_2\angle\theta_2\). The directions are known, so the two lengths follow from a triangle, or from the components:
\[ \begin{aligned} m_3 r_3 \cos\theta_3\; l_3 + m_4 r_4 \cos\theta_4\; l_4 &= -\,m_2 r_2\, l_2 \cos\theta_2 \\ m_3 r_3 \sin\theta_3\; l_3 + m_4 r_4 \sin\theta_4\; l_4 &= -\,m_2 r_2\, l_2 \sin\theta_2 \end{aligned} \]
solved by Cramer's rule. A negative l puts a mass on the other side of mass 1.
Can a solution be set? Every block face must lie on the scale, 0 to 170 mm, and no two blocks may be closer than 20 mm centre to centre, the width of a block. When masses 1 and 2 are fixed, a solution that breaks either rule cannot be set. When they may move, a negative l is no obstacle as long as the whole set spans no more than 170 mm: slide it along the shaft until the lowest face reads 0 or more. Each problem has two solutions. Often only one can be set, and some choices of masses 1 and 2 leave none.
3.7 The rig on its mounts
The rubber mounts let the plate move. Treat the plate, bearings and shaft as one rigid body: vertically it can bounce and pitch (rock about an axis across the shaft), and sideways it can sway and yaw. Each motion has a natural frequency of a few hertz. For a single rotating force \(F = mr\,\omega^2\) acting on a mass M on mounts of total stiffness k, with damping expressed as a loss factor η, the amplitude of the motion is
\[ x = \frac{m r\,\omega^2}{\sqrt{\left(k - M\omega^2\right)^2 + (\eta k)^2}} \]
Well below resonance \(x \approx mr\,\omega^2/k\), so the motion grows as \(\omega^2\) and the velocity as \(\omega^3\). At resonance, \(\omega = \sqrt{k/M}\), damping alone limits it. Well above resonance \(x \to mr/M\): the plate moves like a free mass, and its highest point comes half a turn after the heavy spot passes the top. The lag between the heavy spot and the high spot grows from 0° below resonance through 90° at resonance to 180° above it. A vibrating body moving as \(x \sin\omega t\) has velocity amplitude \(v = \omega x\), and the analyser quotes the RMS value \(v/\sqrt{2}\).
3.8 The analyser, and influence coefficients
OVERALL is the RMS of the whole velocity signal, every frequency included. 1× uses the tacho to pick out only the part at shaft speed and gives its RMS amplitude and its phase: the angle the shaft turns from the tacho pulse to the positive peak of the velocity. The tape is not at 0° on the protractor, so the phase carries an unknown constant offset. Every method on this bench works with differences of readings, so the offset cancels. With this convention, turning a trial block by +30° turns the phase by +30°.
Write each 1× reading as a complex number \(A = \text{amplitude}\angle\text{phase}\), and each block as \(W = mr\angle\theta\). The readings are linear in the unbalance, so at bearing i
\[ A_i = A_{i0} + \alpha_{i1} W_1 + \alpha_{i2} W_2 \]
where \(W_1\) and \(W_2\) are masses added in planes 1 and 2. A trial mass T at 0° in plane 1 alone gives \(\alpha_{11} = (A_1' - A_{10})/T\) and \(\alpha_{21} = (A_2' - A_{20})/T\), and a trial in plane 2 gives \(\alpha_{12}\) and \(\alpha_{22}\). The corrections that cancel the original vibration solve
\[ \alpha_{11} W_1 + \alpha_{12} W_2 = -A_{10}, \qquad \alpha_{21} W_1 + \alpha_{22} W_2 = -A_{20} \]
\[ W_1 = \frac{-A_{10}\,\alpha_{22} + A_{20}\,\alpha_{12}}{\Delta}, \qquad W_2 = \frac{-A_{20}\,\alpha_{11} + A_{10}\,\alpha_{21}}{\Delta}, \qquad \Delta = \alpha_{11}\alpha_{22} - \alpha_{12}\alpha_{21} \]
The blocks on this rig have fixed moments, so replace \(W_1\) and \(W_2\) by the force \(\mathbf{F} = W_1 + W_2\) and the couple \(\mathbf{C} = W_1 z_1 + W_2 z_2\) about the scale zero, where z is a centre plane (face + 10 mm). Then find θ and position for blocks 3 and 4 so that \(m_3r_3\angle\theta_3 + m_4r_4\angle\theta_4 = \mathbf{F}\) and \(m_3r_3 z_3\angle\theta_3 + m_4r_4 z_4\angle\theta_4 = \mathbf{C}\), as in section 3.6. Experiment 7 runs the same equations the other way: the residual unbalance \(U_1, U_2\) that produced the bare-shaft readings solves \(\alpha_{i1}U_1 + \alpha_{i2}U_2 = A_{i0}\).
3.9 The four-run method
Without a phase reference only amplitudes are available. Record the amplitude \(A_0\) as found, then with a trial mass T in the correction plane at 0°, 120° and 240°, giving \(A_1, A_2, A_3\). Draw a circle of radius \(A_0\) about a point O and mark the points at 0°, 120° and 240° on it. Draw circles of radius \(A_1, A_2, A_3\) about those points. They meet at a point S, or close to one. The length OS is the effect of the trial mass alone, and the correction is a mass of moment
\[ m_c r_c = T\,\frac{A_0}{OS} \]
at the angle of OS, measured from the 0° mark in the direction of increasing trial angle. A single plane can cancel the force but not, in general, the couple.
3.10 The stroboscope, the chalk and the coin
A strobe shows the rotor only at its flashes. If the shaft turns a whole number of revolutions between flashes, the image stands still. That happens at the true speed and also at a half, a third and so on of it, so a frozen image alone does not prove the speed. At twice the shaft speed you see two images half a turn apart. A little below the shaft speed the image creeps slowly round in the direction of rotation, and the plate's motion slows with it: slow motion of the real event.
Chalk held lightly against the rim of a running disc marks the high spot, the point that rises to meet it. Below resonance the high spot lies at the heavy spot, and above resonance opposite it (section 3.7).
A coin standing on edge topples when the sideways acceleration of its support exceeds about \(g\,t/D\), with t its thickness and D its diameter: about 1.2 m/s² for a £1 coin. A coin that stays up at full speed is a quick sign of a well-balanced machine.
3.11 Uncertainty and analysis
Resolution and scatter. The analyser resolves 0.1 mm/s and 1°, and noise adds about 0.03 mm/s to each 1× reading. You can read the protractor to about 0.5° and the scale to about 0.5 mm. The caliper resolves 0.02 mm, and the slotted masses are within 0.1 % of their stamped values. Repeat runs, and average them as vectors, to reduce the random part.
Fitting a straight line. For n points the least squares slope and intercept are
\[ m = \frac{\sum (x_i - \bar x)(y_i - \bar y)}{\sum (x_i - \bar x)^2}, \qquad c = \bar y - m\,\bar x, \qquad s_m = \sqrt{\frac{\sum (y_i - m x_i - c)^2}{(n - 2)\sum (x_i - \bar x)^2}} \]
The Plot tab draws this line and reports m, c, their standard errors and R². Carry out the calculations of your results yourself.
Propagating uncertainty. For products and quotients, relative uncertainties combine in quadrature. The moment in g·mm, \(mr = (\text{slope})\times R\), carries \(u(mr)/mr = \sqrt{(u_{\text{slope}}/\text{slope})^2 + (u_R/R)^2}\).
Comparing two estimates. Two values \(x_1 \pm u_1\) and \(x_2 \pm u_2\) are consistent if they differ by less than about twice \(\sqrt{u_1^2 + u_2^2}\). A larger difference points to a systematic effect, which your report should identify.