M-12  Torsion Pendulum and Shear Modulus

OBJECTIVES: To study a torsion pendulum and to measure a shear modulus.

APPARATUS:

Wall mounted torsion pendulum (large disk plus suspension assembly); rods of different lengths, diameters and material; a ring and two cylindrical masses; timer, tally counter.

INTRODUCTION:

If a torque $\tau$ twists a rod, then the angle $\theta$ through which the rod twists is proportional to the torque if the elastic limit is not exceeded: Thus $\tau_{\mbox{\tiny applied}} = \kappa \theta$ where $\kappa$ is the torsion constant.

Figure 1: (a) A thin-walled hollow cylinder. (b) The cylinder is twisted by applying a torque. (c) The angle of shear $\phi$ and the angle of twist $\theta$.

The equal and opposite reaction torque is

$$\tau_{\mbox{\tiny {el}}} \; = \; - \kappa \theta \; = \; I \alpha \; = \; I \frac{d^2\theta}{dt^2} \;.$$

Since this is the same mathematical form as

$$F_{\mbox{\tiny el}} \;= \; -kx \;= \; ma \; = \; m\frac{d^2x}{dt^2},$$

it has a similar solution: $\theta = \theta_0 \sin \omega t$ where $\omega = \sqrt{\kappa/I} = 2 \pi f = 2 \pi/T$. Thus the system undergoes angular simple harmonic motion, SHM, with $T = 2 \pi \sqrt{I/ \kappa}$ and $T$ is independent of amplitude, (in contrast to a simple pendulum where $T$ is only approximately independent of amplitude.) For simple geometries, one readily calculates the rotational inertia $I$. Hence measuring $T$ can determine precisely the torsion constant, $\kappa$. To relate the torsion constant $\kappa$ to the shear modulus $M_s$, consider first a tangential force $dF_t$ applied to the end of a thin ($dr$ thick) hollow tube. This tangential force/area is the shear stress and equals $dF_t/(2\pi r dr$). The resultant shear strain is the angle $\phi$ (see Fig. 1(c)):

$$\phi = \frac{a}{L} = \frac{r\theta}{L}  .$$

Multiplying the stress by $r/r$ gives

$$\frac{rdF_t}{2\pi r^2dr} = \frac{d\tau}{2\pi r^2dr} = \mbox{shear stress.}$$

Hence the shear modulus $M_s$ becomes:

$$M_s = \frac{\mbox{shear stress}}{\mbox{shear strain}} = \fra... ...frac{Ld \tau}{ \theta (2\pi r^3)dr} \hspace{.5in} \mbox{[tube]}$$

and $Ld\tau = M_s \theta(2\pi r^3 dr)$. If one integrates from 0 to $R$, the tube then becomes a solid rod:

$$L \tau = M_s \theta \; 2\pi \left( \frac{R^4}{4}\right) \hspace{.5in} \mbox{[solid rod]}.$$

Thus

$$M_s =\frac{2L}{\pi R^4} \frac{\tau}{\theta} = \left(\frac{2L}{\pi R^4}\right) \kappa  .$$

SUGGESTED EXPERIMENTS: (only one of which need be performed)

1. Determine the shear modulus of one or more materials by measuring the period of a torsion pendulum with the materials as the suspension.
2. Use the result of Exp. 1 (or from a table of shear moduli) to calculate the period of a torsion pendulum with the ring (or the two cylindrical masses) placed on the disc. Compare with the measured value of the period.
3. Check the expression for shear modulus,
   
   $$M_s = \frac{2L\tau}{\pi R^4 \theta} = \frac{2L}{\pi R^4} \kappa .$$
   
   
   where $\kappa = \tau/\theta$ is the torsion constant, R and L are the radius and length of the rod, $\tau$ is the torque and $\theta$ is the angular displacement. Use rods of the same material but different lengths and diameters.
4. Determine how the period of the torsion pendulum depends on the length and diameter of the rod; also on the rotational inertia of the suspended mass. Check that the period is independent of angular amplitude (as long as one is below the elastic limit). Compare with theory.

QUESTION:

Is it reasonable to ignore the rotational inertia of the suspending rod? Check your answer by estimating the ratio of the rod's rotational inertia to that of the large disc.