62 #10 The Torsion Pendulum Objectives To compare the experimentally determined moment of inertia of sev- eral bodies with the corresponding theoretical value. Theory In general, the motion of a rigid body is a combination of translation- al and rotational motion. The description of the rotational portion of the motion involves both rotational kinematics and rotational dynamics. The variables of rotational kinematics are presented in Table 10.1 along with the variables of translational (or rectilinear) motion for the sake of com- parison. Rotation about a Fixed Axis Translational Motion ө Angular displacement Displacement X 0 = d Oldt Angular velocity v = dx/dt Velocity a = d o/dt Angular velocity a = dv/dt Acceleration 1 Moment of inertia Mass M F = Ma Force Torque τ = Ια Work Work W = /Fdx K = 72Mv? W = ſtdo K = 7210 Kinetic energy Kinetic energy Power P = Fy P = To Power Linear momentum P = My L = Io Angular momentum Table 10.1. Correspondence between important quantities for translational and rota- tional motion. Consider a rigid body rotating with angular speed o about an axis that is fixed in a particular reference frame. A particle of mass m at a distance r from the axis of rotation moves in a circle of radius r with an angular speed w about this axis and has a linear speed v = ro. ro. Its kinetic energy is K = } mv = ? 1 mr2 02 2 (10.1) If the body is rigid, w is the same for all particles in the body but r is different for each particle. Hence, the total kinetic energy of the rotating body is K= 1 (min + mori tu.o? = (Emri)o? (10.2) . The term ?mir? is denoted by the symbol I and is called the moment of inertia or the rotational inertia of the body with respect to the particular axis of rotation. Note that the moment of inertia of a body will depend on the axis about which the body rotates as well as on the manner in which the mass is distributed throughout the body. For a body that is not composed of discrete point masses but is a con- tinuous distribution of matter, the equation (10.3) 1 = ?mir? is replaced by (10.4) I = Sr2 dm Table 10.2 at the end of this experiment lists the moments of inertia for a few symmetrical solids. Consider a long uniform wire, rigidly clamped at its upper end and supporting at its lower end a circular disk whose moment of inertia about an axis through the wire is I (Fig. 10.1). If the disk is rotated through some angle, a restoring torque, T, is set up in the wire tending to bring the disk back to its equilibrium position. When the disk is released, this torque gives the disk an angular acceleration (10.5) τ = Τα The restoring torque set up in the wire when its lower end is twisted through some angle 0 (Fig. 10.1) depends upon the torsion constant of the wire as well as the angle of twist. The torsion constant k depends upon the dimensions of the wire and upon the material of which it is made, i.e., up- on the shear modules of the material. This torsion constant k is defined as the torque required to twist the lower end of the wire through unit angle (one radian). Consider the motion of the disk at some instant when the angle of twist is 0 (in radians) and the restoring torque is therefore ko. We must have at this instant ko = -la, or -k Ꮎ a = 1 (10.6) The negative sign is used because a is positive in the direction of increas- ing 0, which is opposite to the direction of the restoring torque ko. Since k and I are constant for a given configuration of the apparatus, it follows that the angular acceleration for an angular displacement is proportional to e. This condition, that the resulting acceleration of a system due to a 64 simple displacement be proportional to the displacement, is the necessary condition for the system to execute simple harmonic motion. Fixed end Suspension wire Reference line + 0 -0% Fig. 10.1. Torsion pendulum displaced an angle 0. Since d20 a = dt2 (10.7) then d²0 kᎾ + I (10.8) 0 = dt? This is a linear homogenous differential equation whose solution is O = Om cos(@t + 0) (10.9) where Om is the angular amplitude, w is the angular frequency, and 0 is a phase angle. Equation 10.9 will be a solution of Eq. 10.8 provided 02 + = 0, +1 -0. (10.10) 65 or k (10.11) 0 VE 2 af VI Solving for the period T (= 1/8) gives T = 24 V (10.12) Note the derivation of the period for the pendulum follows exactly the derivation for the period of the simple pendulum, except that the approxi- mation sin 0 = 0 is not needed. Therefore the amplitude of oscillation should have no effect on the period of the torsion pendulum. Solving Eq. 10.12 first for I and then k yields kT2 I = (10.13) 2 4π and k 4021 T2 (10.14) = Apparatus torsion pendulum (wire and disk) metal ring 2 identical metal cylinders stopwatch meter stick protractor - Experimental Procedure I. Disk 1. Record the mass of the disk as indicated on it. Measure and record its diameter estimate and record the uncertainties in both mass and diame- ter. 2. Set the torsion pendulum into oscillation with an amplitude of about 10°. Record the time for 10 complete oscillations. Repeat this proce- dure two more times. 3. Set the torsion pendulum into oscillation with an amplitude of about 50°. Record the time for 10 complete oscillations. Repeat this proce- dure two more times. 66 II. Disk + Ring 4. Place the ring onto the disk, centering it carefully. Record the mass of the ring as indicated on it. Also, measure and record the inside and out- side diameters of the ring, along with their uncertainties. 5. Set the system into oscillation with an amplitude of about 10° and rec- ord the time for 10 complete oscillations. Repeat two more times. III. Disk + Cylinders 6. Place the ring back up on the support where it was originally. Place the two identical large solid cylinders on the disk so that their index lines are aligned with the diameter line on the disk and the cylinders are tan- gent to the circumference of the disk. The mass of each cylinder is 2.680 0.005 kg. Determine the distance from the axis of rotation to the center of each cylinder. 7. Set this system into oscillation with an amplitude of about 10° and rec- ord the time for 10 complete oscillations. Repeat two more times. Analysis of Data I. Disk 1. For each of the three trials of Procedure step 2 calculate the period T of the torsion pendulum. Find the average period of the three trials along with its uncertainty. Repeat for the data of Procedure step 3. 2. Find the percent difference between the periods of the pendulum with small amplitude and large amplitude. For simple harmonic motion, the period of oscillation is independent of amplitude. Do the two values for the period obtained agree within the uncertainties? Is it to be expected that they will agree? Comment in your Conclusion. 3. Calculate the moment of inertia of the disk and its uncertainty using the formula I disk I even = MR 1 MR2 2 4. Using I from step 3 and the average period T from step 1 above, calcu- late the torsion constant k of the suspension wire from Eq. 10.14 along with its uncertainty. II. Disk + Ring 5. Using the data from Procedure step 5, calculate the average period of the ring plus disk. Use Eq. 10.13 with the value of the torsion constant k found in step 4 above to calculate the combined moment of inertia of the ring plus disk.
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