The photograph above shows two torsion pendulums, each of which has a disc whose mass is 4.57 kg. The torsion wire on the pendulum on the right has a greater diameter than the wire on the pendulum on the left. The torsion wire on the right thus has a greater force constant than that of the one on the left; it is stiffer than the one on the left.
Turn the disc of one of the pendulums in either direction, and then release it. The disc now oscillates back and forth. The pendulum on the right, whose torsion wire is stiffer than that of the one on the left, does so at a higher frequency than does the pendulum on the left. Setting the ring on the disc of either pendulum, and thus increasing the moment of inertia, lowers the frequency of oscillation. Both rings have the same mass, 4.19 kg, and for each pendulum, adding the ring increases the moment of inertia by a factor of about 2.7 (vide infra).
A torsion pendulum is analogous to a mass-spring oscillator. Instead of a mass at the end of a helical spring, which oscillates back and forth along a straight line, hower, it has a mass at the end of a torsion wire, which rotates back and forth. To set the mass-spring in motion, you displace the mass from its equilibrium position by moving it in a straight line and then releasing it. The helical spring (or gravity, depending on whether or not the system is oriented vertically, and in which direction you displace the mass) exerts a (linear) force to restore the mass to its equilibrium position. To set the torsion pendulum oscillating, you turn the mass (rotate it about its center), and then release it. To do this, you must exert a torque about the bottom of the torsion wire. The torsion wire, in turn, exerts a restoring torque to bring the mass back to its original position.
The torsion wires on the pendulums shown in the photograph are brass. The one on the pendulum on the left has a diameter of about 2.8 mm, and the one on the pendulum on the right has a diameter of about 3.9 mm. Both are approximately one meter long. The torsion wire on the pendulum on the right is therefore considerably stiffer than the one on the pendulum on the left.
The disc on each pendulum has a mass of 4.57 kg and a radius of 0.127 m, for a moment of inertia ((1/2)MR2) = 3.69 × 10-2 kg-m2. The rings have a mass of 4.19 kg, and an average radius of (0.122 m), for a moment of inertia (MR2) of 6.24 × 10-2 kg-m2. So adding a ring to a disc increases the moment of inertia by a factor of 2.7 relative to that of the disc alone. A white double arrow on each disc provides a reference by which to observe the oscillatory motion of the pendulum. A white stripe on each ring provides a similar reference. (Make sure that when you set the ring on the disc, the white stripe faces the class.)
When you rotate the disc hanging from the bottom of the torsion wire through some angle θ, the torsion wire exerts a restoring torque, τ = - kθ. This is Hooke’s law, and is analogous to the linear case, F = -kx. For a body subject to a torque, τ = Iα, where I is the body’s moment of inertia, and α is the angular acceleration of the body. From these two equations, we see that -kθ = I(d2θ/dt2). (Ω = dθ/dt, and α = dΩ/dt. Here I use Ω to denote speed of rotation, to avoid confusion with ω, used for frequency of oscillation.) Which we can write -(k/I)θ = (d2θ/dt2), or (d2θ/dt2) + (k/I)θ = 0. This is exactly the same form as the equation for the mass-spring in demonstration 40.12 -- Mass-springs with different spring constants and masses (except for one constant). It thus has the same solution. If we substitute ω = √(k/I), we have
(d2θ/dt2) + ω2θ = 0.
This has the characteristic equation r2 + ω2 = 0, whose roots are r = ±iω. This leads to the general solution
x = C1 cos ωt + C2 sin ωt.
As shown in the page for demonstration 40.12 -- Mass-springs with different spring constants and masses, appropriate substitution for the constants C1 and C2, and use of a trigonometric identity give the solution
θ = θm cos (ωt - φ),
where θm is the maximum displacement of the pendulum from its equilibrium position, and thus the amplitude of the oscillation.
The frequency of oscillation, in units of radians per second, is ω = √(k/I). The frequency in cycles per second, ν, equals ω/(2π). The period of oscillation, T, equals (2π)/ω, or 1/ν. φ is a phase factor, which depends on how the motion starts. We see that the frequency at which the torsional pendulum oscillates is proportional to √k, and inversely proportional to √I.
We also find an interesting relationship among θ, Ω and s, the position, velocity and acceleration of the disc. If we take θ = θm cos (ωt - φ), then Ω = dθ/dt = -ωθm sin (ωt - φ), and α = dθ2/dt2 = -ω2θm cos (ωt - φ). These equations show us that at the extrema (±θm), the angular velocity is zero, and the angular acceleration is at its maximum, and that at the equilibrium position (θ = 0), the angular acceleration is zero and the angular velocity is at its maximum.
The work to displace the disc through a particular angle, and thus the potential energy stored in the torsion wire when the disc is so displaced, is W = ∫τ·dθ which equals ∫kθ dθ, which gives U = (1/2)kθ2. The rotational kinetic energy of the disc is K = (1/2)IΩ2 (where Ω = dθ/dt). The total energy must remain constant, except for frictional losses. The equations above, for θ and Ω, show that at the extrema, the potential energy is at its maximum, and the rotational kinetic energy equals zero. At the equilibrium position, the potential energy equals zero, and the rotational kinetic energy is at its maximum. As noted above, their sum is constant.
References:
1) Thomas, George B., Jr. and Finney, Ross L. Calculus and Analytic
Geometry (Reading, Massachusetts: Addison-Wesley Publishing Company,
1992), pp. 1080-1081.
2) Resnick, Robert and Halliday, David. Physics, Part One, Third
Edition (New York: John Wiley and Sons, 1977), pp. 312.
3)
https://courses.physics.illinois.edu/phys211/su2012/labs/lab7/answer_a.html.