Moment of Inertia Bifilar Suspension Formula

Derivation of the moment of inertia bifilar suspension formula

1. Twist the body through a small angle θ. Each thread's lower end moves sideways by an arc $b\theta$, where b = d/2.
2. Each thread now leans at a small angle φ from the vertical with $L\phi = b\theta$, so $\phi = \frac{b\theta}{L}$.
3. Each thread carries tension about $\frac{mg}{2}$. Its horizontal part, $\frac{mg}{2}\phi$, acts at right angles to the radius b, pulling the body back.
4. Restoring torque from both threads: $\tau = 2 \times \frac{mg}{2} \times \frac{b\theta}{L} \times b = \frac{mgb^2}{L}\theta$.
5. Apply τ = Iα to the twist angle: $I\ddot{\theta} = -\frac{mgb^2}{L}\theta$. This is simple harmonic motion with $\omega^2 = \frac{mgb^2}{IL}$.
6. Period: $T = 2\pi\sqrt{\frac{IL}{mgb^2}}$. Square and rearrange: $I = \frac{mgb^2T^2}{4\pi^2L}$. Put b = d/2 to get $\frac{mgd^2T^2}{16\pi^2L}$.

Running the experiment

1. Hang the body (a bar or a rectangular plate) from two equal threads so it is level. Measure L and d with a metre scale.
2. Find m on a balance.
3. Give a small twist and release. Avoid any swinging motion.
4. Time 20 oscillations, repeat, and divide the mean by 20 to get T.
5. Repeat for several thread lengths. Plot $T^2$ on the y axis and L on the x axis; the points should lie on a straight graph through (0, 0) whose gradient equals $\frac{16\pi^2 I}{mgd^2}$.

Counting twenty swings spreads your stopwatch delay over many cycles, a trick used in any period measurement.

Worked example

Error sources and good practice

• Thread lengths not equal, so the body hangs tilted.
• Swinging as well as twisting, which mixes two motions.
• Large twist angles, which break the small angle step.
• Measuring d between the outsides of the hooks rather than between thread centres.
• Rounding T early; because T is squared, a 1 percent error in T gives a 2 percent error in I.

What changes the twisting period

Read the period form, $T = 2\pi\sqrt{\frac{IL}{mgb^2}}$, as a set of design rules. Longer threads give slower oscillations, since T grows as $\sqrt{L}$. Threads set wider apart give faster ones, because the restoring torque rises with $b^2$. A body whose mass sits far from the vertical axis, such as a bar with weights clipped to its ends, has a larger I and twists more slowly.

Clip two 100 g masses to the ends of the bar in the example and its I rises sharply, so T lengthens. Timing both cases is a neat way to test the parallel axis idea, and it shows why I depends on where mass sits, not just how much there is. Keep the threads, spacing and twist angle the same between runs so that only I changes.

The radius of gyration follows straight from the result, $K = \sqrt{I/m}$, which here is about 0.17 m. More turning results are listed in the rotation and circles section.