# 1. Experiment Title<\/p>\n
The subject of this experiment is **Measurement of Moment of Inertia and Conservation of Angular Momentum**.<\/p>\n
# 2. Objective<\/p>\n
By measuring the angular acceleration of a rotating rigid body, determine the object's **moment of inertia (Moment of Inertia)** experimentally, and understand the dynamics of rotational motion by comparing the experimentally obtained value with the theoretical value calculated from the geometric structure. In addition, experimentally confirm that **angular momentum** is conserved when no external torque acts on the rotating system, and consider the change in energy during this process.<\/p>\n
# 3. Relevant Theory<\/p>\n
## 3.1 Moment of Inertia<\/p>\n
The quantity corresponding to 'mass' in linear motion for a rotating object, representing the tendency of the object to maintain its rotational state. For particles of mass $m_i$ at distances $r_i$ from the axis of rotation, the moment of inertia $I$ is defined as:<\/p>\n
$$I = \\sum m_i r_i^2$$<\/p>\n
For a rigid body with a continuous mass distribution, it is calculated by integrating over infinitesimal mass elements $dm$:<\/p>\n
$$I = \\int r^2 dm$$<\/p>\n
### 3.1.1 Derivation of the Moments of Inertia for a Disk and a Ring<\/p>\n
#### 3.1.1.1 Uniform Solid Disk<\/p>\n
Consider a uniform disk of radius $R$ and total mass $M$ and find its moment of inertia about its central axis. The surface density of the disk is $\\sigma = \\frac{M}{\\pi R^2}$.<\/p>\n
Considering a thin ring at radius $r$ with thickness $dr$, the differential area is $dA = 2\\pi r dr$, so the differential mass is $dm = \\sigma dA = \\frac{M}{\\pi R^2} \\cdot 2\\pi r dr$.<\/p>\n
$$I_{disk} = \\int_0^R r^2 dm = \\int_0^R r^2 \\left( \\frac{2Mr}{R^2} \\right) dr = \\frac{2M}{R^2} \\int_0^R r^3 dr$$<\/p>\n
$$I_{disk} = \\frac{2M}{R^2} \\left[ \\frac{r^4}{4} \\right]_0^R = \\frac{2M}{R^2} \\cdot \\frac{R^4}{4} = \\frac{1}{2}MR^2$$<\/p>\n
#### 3.1.1.2 Thick Ring<\/p>\n
For a ring with inner radius $R_1$, outer radius $R_2$, and mass $M$, the surface density is $\\sigma = \\frac{M}{\\pi(R_2^2 - R_1^2)}$.<\/p>\n
$$I_{ring} = \\int_{R_1}^{R_2} r^2 \\left( \\frac{2Mr}{R_2^2 - R_1^2} \\right) dr = \\frac{2M}{R_2^2 - R_1^2} \\left[ \\frac{r^4}{4} \\right]_{R_1}^{R_2}$$<\/p>\n
$$I_{ring} = \\frac{2M}{R_2^2 - R_1^2} \\cdot \\frac{R_2^4 - R_1^4}{4} = \\frac{M}{2(R_2^2 - R_1^2)}(R_2^2 - R_1^2)(R_2^2 + R_1^2)$$<\/p>\n
$$I_{ring} = \\frac{1}{2}M(R_1^2 + R_2^2)$$<\/p>\n
## 3.2 Derivation of the Relationship between Angular Momentum and Torque (Rotational Form of Newton's Second Law)<\/p>\n
To analyze rotational motion dynamically, convert Newton's second law for linear motion ($F = ma$) into its rotational form.<\/p>\n
For a single particle of mass $m$, Newton's second law in differential form is expressed as the time rate of change of linear momentum ($p = mv$):<\/p>\n
$$F = \\frac{dp}{dt} = m\\frac{dv}{dt}$$<\/p>\n
If the particle has a position vector $r$ from the origin, the torque $\\tau$ acting on the particle is defined as the cross product of the position vector and the force vector:<\/p>\n
$$\\tau = r \\times F$$<\/p>\n
Substituting the differential form of Newton's second law for $F$ gives:<\/p>\n
$$\\tau = r \\times \\frac{dp}{dt} \\tag{1}$$<\/p>\n
The angular momentum $L$ of the particle is defined as the cross product of the position vector and linear momentum:<\/p>\n
$$L = r \\times p$$<\/p>\n
Differentiate this angular momentum with respect to time. By the product rule:<\/p>\n
$$\\frac{dL}{dt} = \\frac{d}{dt}(r \\times p) = \\left( \\frac{dr}{dt} \\times p \\right) + \\left( r \\times \\frac{dp}{dt} \\right)$$<\/p>\n
Here $\\frac{dr}{dt}$ is the particle's velocity $v$, and $p = mv$. The cross product of $v$ and $mv$ is zero ($v \\times mv = 0$) since they are parallel, so the first term vanishes and only the second term remains.<\/p>\n
$$\\frac{dL}{dt} = r \\times \\frac{dp}{dt} \\tag{2}$$<\/p>\n
Comparing (1) and (2) yields the fundamental relation between torque and angular momentum: the net torque acting on a system equals the time rate of change of angular momentum.<\/p>\n
$$\\tau = \\frac{dL}{dt} \\tag{3}$$<\/p>\n
Extend this relation to a rigid body rotating about a fixed axis. When a rigid body rotates with angular velocity $\\omega$, the speed of a particle at radius $r$ is $v = r\\omega$ (in scalar terms), so the particle's angular momentum is:<\/p>\n
$$L = r \\cdot p = r(mv) = mr^2\\omega$$<\/p>\n
Since $mr^2$ is the particle's contribution to the moment of inertia $I$, we have $L = I\\omega$. Substituting into (3) and differentiating with respect to time gives:<\/p>\n
$$\\tau = \\frac{d}{dt}(I\\omega) = I\\frac{d\\omega}{dt} = I\\alpha \\tag{4}$$<\/p>\n
Finally, this yields the rotational analog of $F = m\\frac{dv}{dt}$:<\/p>\n
$$\\tau = r \\times F = I\\alpha = \\frac{dL}{dt}$$<\/p>\n
This equation is the key mathematical basis in this experiment for computing the moment of inertia $I$ from measured rotational acceleration $\\alpha$ of disks and rings.<\/p>\n
## 3.3 Rotational Kinetic Energy and Its Derivation<\/p>\n
Just as the kinetic energy of translational motion is determined by mass and velocity, the kinetic energy of a rigid body rotating about a fixed axis can be defined using the moment of inertia and angular velocity. This rotational kinetic energy can be derived by applying the translational kinetic energy definition to each infinitesimal mass element of the rotating body.<\/p>\n
### 3.3.1 Derivation<\/p>\n
Assume a rigid body rotates about a fixed axis with angular velocity $\\omega$. The body can be considered as composed of many small particles.<\/p>\n
For the $i$-th particle of mass $m_i$ at perpendicular distance $r_i$ from the rotation axis, its tangential speed is:<\/p>\n
$$v_i = r_i \\omega$$<\/p>\n
The translational kinetic energy $K_i$ of this particle is given by Newtonian mechanics as:<\/p>\n
$$K_i = \\frac{1}{2}m_i v_i^2$$<\/p>\n
Substituting $v_i = r_i \\omega$ gives:<\/p>\n
$$K_i = \\frac{1}{2}m_i (r_i \\omega)^2 = \\frac{1}{2} m_i r_i^2 \\omega^2$$<\/p>\n
The total rotational kinetic energy $K_{rot}$ of the rigid body is the sum of the kinetic energies of all particles. For a rigid body, all particles share the same angular velocity $\\omega$, so $\\omega$ can be taken outside the summation:<\/p>\n
$$K_{rot} = \\sum_{i} K_i = \\sum_{i} \\left( \\frac{1}{2} m_i r_i^2 \\omega^2 \\right)$$<\/p>\n
$$K_{rot} = \\frac{1}{2} \\left( \\sum_{i} m_i r_i^2 \\right) \\omega^2$$<\/p>\n
The expression in parentheses $\\sum_{i} m_i r_i^2$ is the moment of inertia $I$ defined in Section 3.1. Replacing it by $I$ yields the final formula for rotational kinetic energy:<\/p>\n
$$K_{rot} = \\frac{1}{2}I\\omega^2$$<\/p>\n
This result is in perfect mathematical analogy with the translational kinetic energy formula $K = \\frac{1}{2}mv^2$: in rotational motion, the role of mass $m$ is played by the moment of inertia $I$, and the role of linear speed $v$ is played by angular speed $\\omega$.<\/p>\n
### 3.3.2 Physical Meaning in This Experiment<\/p>\n
In Experiment B (Conservation of Angular Momentum), dropping a mass ring onto a rotating disk is mechanically equivalent to a perfectly inelastic collision: angular momentum $L$ is conserved because no external torque acts on the system, but due to internal friction the two objects eventually rotate together at the same angular speed. Using the above formulas, one can compute the rotational kinetic energy before and after the collision and show that, although angular momentum is conserved, kinetic energy is dissipated (e.g., to heat) so $\\Delta K_{rot} < 0$.\n\n## 3.4 Conservation of Angular Momentum\n\nIf the net external torque on a system is zero ($\\tau_{ext} = 0$), the total angular momentum of the system remains constant:\n\n$$\\frac{dL}{dt} = 0 \\implies L = I_i \\omega_i = I_f \\omega_f = \\text{Constant}$$\n\nIn this experiment we verify this by changing the system's moment of inertia from $I_i$ to $I_f$ by dropping a ring onto the rotating disk and observing the change in angular velocity from $\\omega_i$ to $\\omega_f$.\n\n## 3.5 Experimental Principle for Measuring Moment of Inertia and Formula Derivation\n\nIn this experiment, a string is wound around an axle (radius $r$) and a hanging mass $m$ is attached to the end; the mass is allowed to fall freely. As the mass falls under gravity it pulls on the string, producing a tension $T$ that exerts a torque on the axle and causes the system to rotate. The mechanics can be expressed as follows.\n\n### 3.5.1 Translational Equation of Motion for the Hanging Mass\n\nThe net force on the falling mass $m$ is gravity $mg$ downward and the tension $T$ upward. Taking the downward direction as positive, Newton's second law for the linear acceleration $a$ of the mass is:\n\n$$mg - T = ma \\tag{1}$$\n\n### 3.5.2 Rotational Equation of Motion for the Rigid Body\n\nThe tension $T$ applied tangentially at axle radius $r$ produces a net torque $\\tau$. If the total moment of inertia of the rotating assembly is $I$ and the angular acceleration is $\\alpha$, the torque equation is (the tension and radius are perpendicular so $\\sin 90^\\circ = 1$):\n\n$$\\tau = r \\times T = rT = I\\alpha \\tag{2}$$\n\n### 3.5.3 Kinematic Constraint between Linear and Angular Acceleration\n\nAssuming the string does not slip or stretch, the linear acceleration $a$ of the hanging mass and the tangential acceleration at the axle surface are identical, so:\n\n$$a = r\\alpha \\tag{3}$$\n\n### 3.5.4 Algebraic Derivation of the Final Formula\n\nEliminating variables using the three equations above yields an expression for the moment of inertia $I$ in terms of measurable quantities.\n\nSubstitute $a = r\\alpha$ from (3) into (1) and solve for $T$:\n\n$$mg - T = m(r\\alpha)$$\n\n$$T = m(g - r\\alpha) \\tag{4}$$\n\nSubstitute this tension into the rotational equation (2):\n\n$$r \\cdot \\left[ m(g - r\\alpha) \\right] = I\\alpha$$\n\nExpanding the left side gives:\n\n$$mgr - mr^2\\alpha = I\\alpha$$\n\nDivide both sides by $\\alpha$ to solve for $I$:\n\n$$I = \\frac{mgr - mr^2\\alpha}{\\alpha} = \\frac{mgr}{\\alpha} - mr^2$$\n\nFactoring out $mr^2$ yields the final experimental formula for the moment of inertia:\n\n$$I = mr^2 \\left( \\frac{g}{r\\alpha} - 1 \\right)$$\n\nUsing this formula, we can determine the moment of inertia $I$ by plugging in directly measured geometric constants (hanging mass $m$, axle radius $r$), gravitational acceleration $g$, and the angular acceleration $\\alpha$ obtained via linear regression from SPARKvue data collection.\n\n# 4. Experimental Procedure\n\nThe experiment is carried out in two main parts. In **Experiment A** the falling mass method is used to measure the moments of inertia of the disk and ring, and in **Experiment B** a ring is dropped onto a rotating disk to verify conservation of angular momentum.\n\n## 4.1 Experiment A: Measurement of Moment of Inertia\n\n<\/p>\n
1. **Set up the rotational apparatus**: Assemble the rotational stand and attach the disk to the axle as shown in [Figure 3]. Use a level to ensure the rotational stand and disk are perfectly horizontal.<\/p>\n
2. **Set up the Smart Gate**: Secure the Smart Gate to the rotational stand and adjust its position so that the sensor can accurately detect the groove (spoke) on the pulley attached to the axle.<\/p>\n
3. **Measure basic parameters**: Measure the total mass of the hanging mass and hanger $m$ using an electronic scale, and measure the axle radius $r$ where the string winds using Vernier calipers with precision.<\/p>\n
4. **Prepare the software**: Run the SPARKvue app and select **[Smart Gate Only]** -> [Smart Pulley (Rotational)]. Set the **Spoke Angle** to $36^\\circ$ (or the appropriate value for the apparatus) and choose **Velocity** and **Acceleration** as the measured quantities.<\/p>\n
5. **Measure the disk's angular acceleration**: With the string wound on the axle, release the hanging mass and record the angular velocity during the fall. Select an interval in the data where the angular acceleration $\\alpha$ is approximately constant and perform a linear fit to obtain the slope.<\/p>\n
<\/p>\n
6. **Repeat measurements**: Repeat step 5 five times to obtain the average angular acceleration for the disk and use it to calculate $I_{disk}$.<\/p>\n
7. **Add the mass ring**: Place the ring on the disk and repeat steps 5\u20136. The measured value now corresponds to the combined moment of inertia $I_{total}$; subtract the previously obtained $I_{disk}$ to obtain the experimental value for $I_{ring}$.<\/p>\n
## 4.2 Experiment B: Conservation of Angular Momentum<\/p>\n
1. **Reconfigure the apparatus**: Remove the string and hanging mass used in Experiment A. The system should be free to rotate with no external torque.<\/p>\n
2. **Start rotation and measurement**: Spin the disk by hand to a stable rotation and start recording in SPARKvue.<\/p>\n
3. **Measure initial angular velocity ($\\omega_1$)**: Record the angular velocity $\\omega_1$ just before dropping the ring.<\/p>\n
4. **Drop the ring**: Carefully drop the ring onto the rotating disk, ensuring it seats correctly at the disk center.<\/p>\n
5. **Measure final angular velocity ($\\omega_2$)**: After the ring has settled and the disk and ring rotate together, record the stable angular velocity $\\omega_2$.<\/p>\n
6. **Data analysis**: Using the measured $\\omega_1, \\omega_2$ and the $I_{disk}, I_{total}$ from Experiment A, compute the angular momenta $L_1, L_2$ before and after the collision to check for conservation.<\/p>\n
7. **Repeat measurements**: Repeat steps 2\u20136 five times to ensure data reliability.<\/p>\n
# 5. Results<\/p>\n
## 5.1 Experimental Constants<\/p>\n
The geometric constants and measured masses used in calculations for Experiments A and B are as follows.<\/p>\n
Parameter<\/strong><\/th>\n| Symbol<\/strong><\/th>\n | Value<\/strong><\/th>\n | Unit<\/strong><\/th>\n<\/tr>\n<\/thead>\n\n | Hanging Mass<\/strong><\/td>\n | $m$<\/td>\n | 0.145<\/td>\n | $\\text{kg}$<\/td>\n<\/tr>\n | Axle Radius<\/strong><\/td>\n | $r$<\/td>\n | 0.0115<\/td>\n | $\\text{m}$<\/td>\n<\/tr>\n | Disk Mass<\/strong><\/td>\n | $M_{disk}$<\/td>\n | 1.427<\/td>\n | $\\text{kg}$<\/td>\n<\/tr>\n | Disk Radius<\/strong><\/td>\n | $R_{disk}$<\/td>\n | 0.1145<\/td>\n | $\\text{m}$<\/td>\n<\/tr>\n | Ring Mass<\/strong><\/td>\n | $M_{ring}$<\/td>\n | 1.441<\/td>\n | $\\text{kg}$<\/td>\n<\/tr>\n | Ring Inner Radius<\/strong><\/td>\n | $R_1$<\/td>\n | 0.054<\/td>\n | $\\text{m}$<\/td>\n<\/tr>\n | Ring Outer Radius<\/strong><\/td>\n | $R_2$<\/td>\n | 0.0635<\/td>\n | $\\text{m}$<\/td>\n<\/tr>\n | Gravity<\/strong><\/td>\n | $g$<\/td>\n | 9.8<\/td>\n | $\\text{m\/s}^2$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n | ## 5.2 Experiment A: Measurement of Rotational Inertia<\/p>\n ### 5.2.1 Measurement Data and Calculations<\/p>\n We measured the angular acceleration $\\alpha$ by releasing the hanging mass five times for the disk alone and for the disk with the ring. The experimental moment of inertia was calculated using the formula $I = mr^2 (\\frac{g}{r\\alpha} - 1)$. The ring-only moment of inertia was obtained by subtracting the disk's moment from the combined moment ($I_{ring} = I_{total} - I_{disk}$).<\/p>\n
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![\"\"](\"https:\/\/wuhanqing.cn\/resource\/EXP05_IMG\/ExpA_Disk_Combo_Chart.png\")
<\/p>\n![\"\"](\"https:\/\/wuhanqing.cn\/resource\/EXP05_IMG\/ExpA_Ring_Combo_Chart.png\")
<\/p>\n![\"\"](\"https:\/\/wuhanqing.cn\/resource\/EXP05_IMG\/ExpB_Momentum_Combo_Chart.png\")
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