Rotational Dynamics: Torque, Equilibrium, and Moment of Inertia

Fundamentals of Rotational Dynamics and Torque

• Definition of Rotational Dynamics:

  • Rotational dynamics is the study of the causes of rotational motion and the relationships between torques, moments of inertia, and angular accelerations.

  • While linear dynamics investigates how a net force creates linear acceleration ($F_{\text{net}} = m a$), rotational dynamics examines how a net torque causes angular acceleration ($\tau_{\text{net}} = I \alpha$).

• Force in Linear vs. Rotational Systems:

  • In linear motion, a force ($F$) acting on a free body causes translation along a straight path.

  • In rotational motion, an applied force acts upon a rigid body constrained to rotate around a fixed pivot or axis of rotation, causing rotational displacement.

• Definition of Torque ($\tau$):

  • Torque (symbol: $\tau$, Greek letter tau) is the quantitative measure of the effectiveness of a force in producing or altering the rotational motion of an object about a designated pivot or axis.

  • SI Unit of Torque: Newton-meter ($\text{N}\cdot\text{m}$.

  • Torque is a vector quantity having both magnitude and rotational direction (sense of rotation).

• Physical Factors Influencing Torque:

  • Magnitude of the Applied Force ($F$): Torque is directly proportional to the magnitude of the applied force. A larger force creates a correspondingly larger rotational effect.

  • Radial Distance ($r$): Torque is directly proportional to the distance from the pivot or axis of rotation to the point of force application. Applying the same force further away from the axis produces significantly greater torque.

  • Angle of Application ($\theta$): Torque depends critically on the angle $\theta$ between the direction of the applied force and the radial line connecting the pivot to the application point.

• Real-World Demonstration: The Door and Doorknob Analogy:

  • Optimal Placement of Doorknobs: Doorknobs are positioned on the outer edge opposite the hinges to maximize the distance ($r$) from the axis of rotation (the hinges). This allows a person to generate sufficient torque to open the door using minimal force.

  • Pulling at Varying Positions:

  • Pushing or pulling directly at the hinge line ($r = 0$) produces zero torque ($\tau = 0$), making it impossible to rotate the door regardless of force magnitude.

  • Pushing near the middle requires roughly double the force to generate the same torque as pushing at the edge.

  • Pulling at Varying Angles:

  • Pulling perpendicularly ($\theta = 90^\circ$) maximizes torque because $\sin(90^\circ) = 1$.

  • Pulling obliquely ($\theta < 90^\circ$) diminishes the effective rotational force component.

  • Pulling parallel to the door face toward or away from the hinges ($\theta = 0^\circ$ or $\theta = 180^\circ$) yields zero torque ($\sin(0^\circ) = \sin(180^\circ) = 0$), producing only linear tension or compression with no rotation.

Mathematical Formulation of Torque and the Lever Arm

• The General Torque Equation:

  • Torque is mathematically defined as:     $\tau = F r \sin(\theta)$

  • $F$: Magnitude of the applied force in Newtons ($\text{N}$

  • $r$: Radial distance from the pivot/axis of rotation to the point where the force is exerted in meters ($\text{m}$

  • $\theta$: Angle measured between the position vector $\mathbf{r}$ and the force vector $\mathbf{F}$

• Sign Conventions for Torque:

  • Counterclockwise (CCW) Rotation: Assigned a positive sign ($+\tau$).

  • Clockwise (CW) Rotation: Assigned a negative sign ($-\tau$).

• The Concept of the Lever Arm (Force Arm, $L$):

  • Definition: The lever arm ($L$) is the perpendicular distance from the axis of rotation to the line of action of the applied force.

  • Mathematical Formula:     $L = r \sin(\theta)$

  • Alternative Expression for Torque:     $\tau = F L$

  • Critical Angle Scenarios:

  • When the force is applied perpendicularly ($\theta = 90^\circ$):       $L = r \sin(90^\circ) = r$       $\tau = F r$

  • When the force is directed at an oblique angle ($0^\circ < \theta < 90^\circ$):       $L = r \sin(\theta) < r$       The effective lever arm decreases, reducing the torque produced.

  • When the line of action passes directly through the axis of rotation ($\theta = 0^\circ$ or $180^\circ$):       $L = r \sin(0^\circ) = 0$       $\tau = 0$

• Mechanical Advantage of Wrench Lengths:

  • A short wrench provides a small radial distance ($r$), requiring a large applied force ($F$) to reach the necessary torque.

  • A long wrench provides a large radial distance ($r$), requiring a much smaller applied force ($F$) to produce identical torque.

  • Increasing the lever arm amplifies the rotational mechanical advantage, making stubborn fasteners easier to turn.

Equilibrium of Torques and Balanced Systems

• Balanced vs. Unbalanced Linear Systems (Review):

  • Balanced Forces ($F_{\text{net}} = 0\,\text{N}$):

  • The object experiences no linear acceleration ($a = 0\,\text{m/s}^2$).

  • The object is either permanently at rest ($v = 0\,\text{m/s}$) or translating at a constant velocity ($v = \text{constant}$).

  • Unbalanced Forces ($F_{\text{net}} \neq 0\,\text{N}$):

  • The object accelerates in the direction of the net force according to Newton's Second Law:       $a = \frac{F_{\text{net}}}{m}$

• Balanced vs. Unbalanced Rotational Systems:

  • Balanced Torques (Rotational Equilibrium, $\tau_{\text{net}} = 0\,\text{N}\cdot\text{m}$):

  • The sum of all clockwise torques exactly counterbalances the sum of all counterclockwise torques:       $\sum \tau = 0$       $\tau_{\text{ccw}} = \tau_{\text{cw}}$

  • The object experiences zero angular acceleration ($\alpha = 0\,\text{rad/s}^2$).

  • The object remains static (at rest, not rotating) or continues rotating at a constant angular velocity ($\omega = \text{constant}$).

  • Unbalanced Torques ($\tau_{\text{net}} \neq 0\,\text{N}\cdot\text{m}$):

  • The object undergoes angular acceleration ($\alpha$) in the direction of the resultant net torque:       $\alpha = \frac{\tau_{\text{net}}}{I}$

• General Equilibrium Condition for a Two-Mass Lever/Seesaw:

  • Consider two masses, $m_1$ and $m_2$, suspended at distances $r_1$ and $r_2$ on opposite sides of a fulcrum:     $\tau_1 = F_{g1} r_1 \sin(90^\circ) = m_1 g r_1$     $\tau_2 = F_{g2} r_2 \sin(90^\circ) = m_2 g r_2$

  • For the system to remain balanced without rotating:     $\tau_{\text{net}} = \tau_1 - \tau_2 = 0$     $m_1 g r_1 = m_2 g r_2$

  • Dividing both sides by the acceleration due to gravity ($g$) yields the mass-distance balance equation:     $m_1 r_1 = m_2 r_2$

Net Torque and Rotational Dynamics Problem Solving

• Step-by-Step Procedure for Calculating Net Torque ($\tau_{\text{net}}$):

  • Step 1: Identify the Axis of Rotation: Clearly locate the pivot point about which torques are being evaluated.

  • Step 2: Determine Each Force and its Geometry: For every individual force $F_i$, measure the radial distance $r_i$ from the axis to the force application point and find the angle $\theta_i$ between $\mathbf{r}_i$ and $\mathbf{F}_i$.

  • Step 3: Calculate Individual Torques: Evaluate $\tau_i = F_i r_i \sin(\theta_i)$.

  • Step 4: Assign Directions (Signs): Assign a positive (+$) value to counterclockwise torques and a negative (-) value to clockwise torques.\n - **Step 5: Compute the Algebraic Sum:**\n    \tau_{\text{net}} = \sum \tau_i = \tau_1 + \tau_2 + \tau_3 + \dots\n - **Step 6: Interpret the Result:**\n - If \tau_{\text{net}} > 0, the system accelerates counterclockwise.\n - If \tau_{\text{net}} < 0, the system accelerates clockwise.\n - If \tau_{\text{net}} = 0, the system is in complete rotational equilibrium.\n\n- **Multi-Force System Example: Merry-Go-Round Wheel:**\n - Consider a circular wheel with two workers applying tangential forces:\n - Worker 1 applies F_1 = 100\,\text{N}$at radius$R_1 = 2\,\text{m} directed tangentially counterclockwise:\n      \tau_1 = + (100\,\text{N})(2\,\text{m}) = +200\,\text{N}\cdot\text{m}\n - Worker 2 applies F_2 = 200\,\text{N}$at radius$R_2 = 1\,\text{m} directed tangentially counterclockwise:\n      \tau_2 = + (200\,\text{N})(1\,\text{m}) = +200\,\text{N}\cdot\text{m}\n - Total Net Torque:\n      \tau_{\text{net}} = \tau_1 + \tau_2 = +200\,\text{N}\cdot\text{m} + 200\,\text{N}\cdot\text{m} = +400\,\text{N}\cdot\text{m}\text{ (counterclockwise)}\n\n\n\n# Moment of Inertia and Rotational Inertia\n\n- **Definition of Moment of Inertia (I):**\n - Moment of inertia (I) represents the rotational counterpart of mass; it measures the resistance of an object to changes in its rotational state of motion.\n - In linear motion, mass (m$) resists changes in linear velocity ($a = F / m).\n - In rotational motion, moment of inertia (I$) resists changes in angular velocity ($\alpha = \tau / I).\n - SI Unit of Moment of Inertia: kilogram-meter squared (\text{kg}\cdot\text{m}^2).\n\n- **The Ice Skater Demonstration:**\n - When an ice skater extends their arms and leg outward, their mass is distributed farther from the axis of rotation, resulting in a large moment of inertia (I_{\text{large}}$) and a low spinning speed ($\omega_{\text{small}}).\n - When the skater draws their arms tightly inward to the chest, their mass becomes concentrated very close to the axis of rotation, drastically reducing their moment of inertia (I_{\text{small}}$) and producing a dramatic increase in spinning speed ($\omega_{\text{large}}).\n - This demonstrates that rotational inertia depends not simply on total mass, but on how that mass is spatially distributed relative to the axis.\n\n\n\n- **Moment of Inertia of a Point Mass:**\n - For a single point mass (m$) revolving at a distance ($r) from a fixed axis:\n    I = m r^2\n - m$: Mass of the particle in kilograms ($\text{kg}\n - r$: Perpendicular distance from the axis of rotation in meters ($\text{m}\n\n- **Moment of Inertia of a Discrete System of Particles:**\n - For a system consisting of multiple point masses rigidly connected:\n    I = \sum_{i=1}^n m_i r_i^2 = m_1 r_1^2 + m_2 r_2^2 + \dots + m_n r_n^2\n\n- **Factors Determining the Moment of Inertia of Extended Objects:**\n - **Total Mass (m):** Larger total mass scales the moment of inertia upward.\n - **Geometry and Dimensions:** Length, radius, or width directly influence r^2.\n - **Mass Distribution:** The more mass located far from the axis of rotation, the larger the moment of inertia.\n - **Location and Orientation of the Axis of Rotation:** An identical physical object has completely different moments of inertia depending on where the axis is placed.\n\n# Moment of Inertia of Standard Extended Bodies\n\n- **General Formula Structure:**\n - For uniform geometric objects of mass m$and characteristic radius$r$(or length$l):\n    I = c m r^2\n - Here, c$is a dimensionless geometric constant ($c \le 1) dictated by the mass distribution relative to the axis.\n\n- **Table of Standard Moments of Inertia:**\n\n  | Object Type | Axis Location | Diagrammatic Representation | Formula for Moment of Inertia (I) |\n  | :--- | :--- | :--- | :--- |\n  | **Thin hoop / ring** of radius r$| Through central perpendicular axis | Ring with all mass at outer perimeter |$I = m r^2$($c = 1) |\n  | **Solid uniform cylinder / disk** of radius r$| Through central symmetry axis | Solid cylinder with mass spread from center |$I = \frac{1}{2} m r^2 = 0.5 m r^2 |\n  | **Uniform solid sphere** of radius r$| Through center | Solid sphere with mass concentrated near center |$I = \frac{2}{5} m r^2 = 0.4 m r^2 |\n  | **Long uniform rod** of length l$| Through center, perpendicular to length | Symmetric rod about midpoint |$I = \frac{1}{12} m l^2 |\n  | **Long uniform rod** of length l$| Through end, perpendicular to length | Pivoted rod swinging about one extremity |$I = \frac{1}{3} m l^2 |\n  | **Thin rectangular plate** of length l$, width$w$| Through center, perpendicular to plate | Flat sheet rotating about centroid |$I = \frac{1}{12} m (l^2 + w^2) |\n\n\n\n- **The Incline Rolling Race Phenomenon:**\n - When various objects of identical mass and radius roll down an incline without slipping:\n - A frictionless sliding block converts all potential energy directly into linear translation, accelerating fastest.\n - For rolling objects, potential energy must be partitioned into both translational kinetic energy (\frac{1}{2} m v^2$) and rotational kinetic energy ($\frac{1}{2} I \omega^2).\n - Objects with a smaller constant c retain less rotational resistance and allocate more energy to translational velocity, reaching the bottom faster.\n - **Race Order (Fastest to Slowest):**\n 1. Solid Sphere (c = 0.4\n 2. Solid Cylinder / Disk (c = 0.5\n 3. Spherical Shell (c = \frac{2}{3} \approx 0.67\n 4. Thin Hoop / Cylindrical Shell (c = 1.0) — rolls down the slowest.\n\n\n\n- **Mass Distribution Comparison for Cylindrical Geometries:**\n - Consider five cylinders of identical total mass M$and outer radius$R:\n - Solid cylinder: Mass distributed smoothly from center to perimeter (I is lowest).\n - Thick-walled hollow cylinders: Mass pushed outward (I increases as inner void grows).\n - Thin-walled cylindrical shell / hoop: All mass concentrated at maximum distance R$($I = M R^2 is highest).\n - Therefore, ranking from least to greatest moment of inertia places the solid cylinder first and the thin hoop last.\n\n# Comprehensive Worked Examples and Practice Problems\n\n- **Problem 1: Door with Six Applied Forces (Zero Torque Identification):**\n - *Scenario:* Six forces act on a door mounted on hinges.\n - *Conditions for Zero Torque:* \tau = F r \sin(\theta) = 0\n - Either the radial distance is zero (r = 0), meaning force is applied directly along the hinge axis.\n - Or the angle is 0^\circ$or$180^\circ$($\sin(\theta) = 0), meaning the line of action points straight at the hinge.\n - Forces directed along the plane of the door pointing straight into the hinges produce no rotational effect (\tau = 0).\n\n- **Problem 2: Wooden Board with Greatest Torque:**\n - *Scenario:* Comparing forces applied to a wooden board pivoted at one end:\n - Force at pivot (r = 0\implies \tau = 0\n - Force along board axis toward pivot (\theta = 180^\circ \implies \tau = 0\n - Force upward at the midpoint (r = L/2, \theta = 90^\circ \implies \tau = 0.5 F L\n - Force upward at the far end (r = L, \theta = 90^\circ \implies \tau = F L\n - *Conclusion:* The greatest torque occurs when the maximum perpendicular force is applied at the furthest distance from the pivot (at the extreme end).\n\n- **Problem 3: Minimum Angle on a Wrench:**\n - *Scenario:* A wrench has length r = 0.234\,\text{m}$. A job requires$\tau = 32.4\,\text{N}\cdot\text{m}.\n - *Part a:* An applied force of F = 232\,\text{N}.\n    \tau = F r \sin(\theta)\n    \sin(\theta) = \frac{\tau}{F r} = \frac{32.4\,\text{N}\cdot\text{m}}{(232\,\text{N})(0.234\,\text{m})} = \frac{32.4}{54.288} \approx 0.5968\n    \theta = \arcsin(0.5968) \approx 36.6^\circ\n - *Part b:* A friend exerts F = 275\,\text{N}.\n    \sin(\theta) = \frac{32.4\,\text{N}\cdot\text{m}}{(275\,\text{N})(0.234\,\text{m})} = \frac{32.4}{64.35} \approx 0.5035\n    \theta = \arcsin(0.5035) \approx 30.2^\circ\n\n- **Problem 4: Calculating Force for a Required Torque:**\n - *Scenario:* A wrench of length 24\,\text{cm} = 0.24\,\text{m}$applies torque at an angle$\theta = 90^\circ$. Torque produced is$\tau = 16\,\text{N}\cdot\text{m}.\n    F = \frac{\tau}{r \sin(\theta)} = \frac{16\,\text{N}\cdot\text{m}}{(0.24\,\text{m})(\sin(90^\circ))} = \frac{16}{0.24} \approx 66.7\,\text{N}\n\n- **Problem 5: Angled Force on an Adjustable Wrench:**\n - *Scenario:* Force F = 120\,\text{N}$applied at$r = 0.3\,\text{m}$with an angle$\theta = 115^\circ.\n    \tau = F r \sin(\theta) = (120\,\text{N})(0.3\,\text{m})\sin(115^\circ) = (36)(0.9063) \approx 32.6\,\text{N}\cdot\text{m}\n\n- **Problem 6: Screwdriver Lever Arm Calculation:**\n - *Scenario:* Force F = 5\,\text{N}$acts on a screwdriver at$r = 13\,\text{cm} = 0.13\,\text{m}$at an angle of$\theta = 30^\circ.\n - *Lever Arm:* L = r \sin(\theta) = (0.13\,\text{m})\sin(30^\circ) = (0.13)(0.5) = 0.065\,\text{m} = 6.5\,\text{cm}\n - *Torque:* \tau = F L = (5\,\text{N})(0.065\,\text{m}) = 0.325\,\text{N}\cdot\text{m}\n\n- **Problem 7: Net Torque on a Seesaw (Pivoted at Center C):**\n - *Given Forces:*\n - F_1 = 2\,\text{N}$downward at$r_1 = 4\,\text{m} to the left (tends to rotate CCW, positive):\n      \tau_1 = + (2\,\text{N})(4\,\text{m}) = +8\,\text{N}\cdot\text{m}\n - F_2 = 3\,\text{N}$upward at$r_2 = 2\,\text{m} to the left (tends to rotate CW, negative):\n      \tau_2 = - (3\,\text{N})(2\,\text{m}) = -6\,\text{N}\cdot\text{m}\n - F_3 = 4\,\text{N}$upward at$r_3 = 2\,\text{m} to the right (tends to rotate CCW, positive):\n      \tau_3 = + (4\,\text{N})(2\,\text{m}) = +8\,\text{N}\cdot\text{m}\n - *Resultant Net Torque:*\n    \tau_{\text{net}} = +8\,\text{N}\cdot\text{m} - 6\,\text{N}\cdot\text{m} + 8\,\text{N}\cdot\text{m} = +10\,\text{N}\cdot\text{m}\n - The net torque is 10\,\text{N}\cdot\text{m} in the counterclockwise direction.\n\n- **Problem 8: Equilibrium of Fixed Concentric Pulleys:**\n - *Scenario:* Two rigidly connected pulleys rotate about a shared axis. The right bucket (m_2 = 5.0\,\text{kg}$) hangs at radius$r_2 = 1.2\,\text{cm}$. The left bucket ($M$) hangs at radius$r_1 = 6.0\,\text{cm}.\n\n\n\n - *Equilibrium Condition:*\n    \tau_1 = \tau_2\n    M g r_1 = m_2 g r_2\n    M r_1 = m_2 r_2\n    M (6.0\,\text{cm}) = (5.0\,\text{kg})(1.2\,\text{cm})\n    M = \frac{6.0}{6.0} = 1.0\,\text{kg}\n\n- **Problem 9: Balancing Fruit Baskets:**\n - *Scenario:* Left basket of unknown mass m$hangs at$r_1 = 0.28\,\text{m}$. Right basket of mass$0.8\,\text{kg}$hangs at$r_2 = 0.16\,\text{m}.\n\n\n\n - *Calculation:*\n    m r_1 = m_2 r_2\n    m (0.28\,\text{m}) = (0.8\,\text{kg})(0.16\,\text{m}) = 0.128\n    m = \frac{0.128}{0.28} \approx 0.457\,\text{kg} \approx 0.46\,\text{kg}\n\n- **Problem 10: Fatimah and Aisha on a Seesaw:**\n - *Scenario:* Fatimah has mass m_F = 56\,\text{kg}$. Aisha has mass$m_A = 43\,\text{kg}$and sits at distance$r_A = 1.5\,\text{m} from the pivot.\n - *Distance for Balance (d):*\n    m_F d = m_A r_A\n    (56\,\text{kg}) d = (43\,\text{kg})(1.5\,\text{m}) = 64.5\n    d = \frac{64.5}{56} \approx 1.15\,\text{m}\n\n- **Problem 11: Kariann and Aysha Pivot Placement (Massless 1.75 m Seesaw):**\n - *Scenario:* Kariann (m_K = 56\,\text{kg}$) and Aysha ($m_A = 43\,\text{kg}$) sit at opposite ends of a$1.75\,\text{m}$seesaw. Find distance$r_K from Kariann to the pivot.\n - *Relationship between distances:* r_A = 1.75\,\text{m} - r_K\n - *Balancing Torques:*\n    m_K r_K = m_A r_A\n    56 r_K = 43 (1.75 - r_K)\n    56 r_K = 75.25 - 43 r_K\n    99 r_K = 75.25\n    r_K = \frac{75.25}{99} \approx 0.76\,\text{m}\n - The pivot point must be placed 0.76\,\text{m}$from Kariann (or$0.99\,\text{m} from Aysha).\n\n- **Problem 12: Balancing a Non-Uniform/Off-Center Pivoted Beam with Weight:**\n - *Scenario:* A beam is 2.0\,\text{m}$long and weighs$20\,\text{N}$. Its center of gravity is at its geometric center ($1.0\,\text{m}$from the left end). It is pivoted at a point$0.5\,\text{m}$from the left end. A downward force$F$acts at the left end ($0.5\,\text{m}$from pivot). A downward force of$10\,\text{N}$acts at the right end ($1.5\,\text{m} from pivot).\n - *Torque Analysis about Pivot:*\n - Force F creates counterclockwise torque:\n      \tau_{\text{ccw}} = F (0.5\,\text{m})\n - Beam weight (20\,\text{N}$) acts at center of mass ($1.0\,\text{m} - 0.5\,\text{m} = 0.5\,\text{m} to the right of pivot, clockwise):\n      \tau_{\text{weight}} = (20\,\text{N})(0.5\,\text{m}) = 10\,\text{N}\cdot\text{m}\n - Right-end force (10\,\text{N}$) acts at$1.5\,\text{m} to the right of pivot (clockwise):\n      \tau_{\text{right}} = (10\,\text{N})(1.5\,\text{m}) = 15\,\text{N}\cdot\text{m}\n - *Sum of Torques = 0:*\n    F (0.5\,\text{m}) = 10\,\text{N}\cdot\text{m} + 15\,\text{N}\cdot\text{m} = 25\,\text{N}\cdot\text{m}\n    F = \frac{25\,\text{N}\cdot\text{m}}{0.5\,\text{m}} = 50\,\text{N}\n\n- **Problem 13: Mass of a Solid Sphere from Moment of Inertia:**\n - *Scenario:* A uniform solid sphere has radius r = 2.0\,\text{m}$and moment of inertia$I = 32\,\text{kg}\cdot\text{m}^2.\n    I = \frac{2}{5} m r^2\n    32 = \frac{2}{5} m (2.0)^2 = \frac{2}{5} m (4.0) = 1.6 m\n    m = \frac{32}{1.6} = 20\,\text{kg}\n\n- **Problem 14: Baton Model (Twirling Baton):**\n - *Scenario:* A massless thin rod of length 0.66\,\text{m}$connects two round end masses of$0.30\,\text{kg} each.\n - *Case 1: Axis through Midpoint:*\n - Radial distance to each mass: r = \frac{0.66\,\text{m}}{2} = 0.33\,\text{m}\n      I_{\text{mid}} = m_1 r_1^2 + m_2 r_2^2 = 2 (0.30\,\text{kg})(0.33\,\text{m})^2 = 2 (0.30)(0.1089) = 0.06534\,\text{kg}\cdot\text{m}^2 \approx 0.065\,\text{kg}\cdot\text{m}^2\n - *Case 2: Axis through One End:*\n - First mass is on the axis (r_1 = 0\,\text{m}).\n - Second mass is at the far end (r_2 = 0.66\,\text{m}).\n      I_{\text{end}} = (0.30\,\text{kg})(0\,\text{m})^2 + (0.30\,\text{kg})(0.66\,\text{m})^2 = (0.30)(0.4356) = 0.13068\,\text{kg}\cdot\text{m}^2 \approx 0.131\,\text{kg}\cdot\text{m}^2\n - *Comparison:* I_{\text{end}} = 2 I_{\text{mid}}. The moment of inertia is twice as large when rotated about one end.\n\n- **Problem 15: Two Balls on a Light Rod:**\n - *Scenario:* Two balls of mass 0.45\,\text{kg}$each are attached to the ends of a light rod of length$0.46\,\text{m}$. Find$I about the center.\n    r = \frac{0.46\,\text{m}}{2} = 0.23\,\text{m}\n    I = 2 m r^2 = 2 (0.45\,\text{kg})(0.23\,\text{m})^2 = (0.90)(0.0529) = 0.04761\,\text{kg}\cdot\text{m}^2 \approx 0.048\,\text{kg}\cdot\text{m}^2\n\n- **Problem 16: Thin Hoop Moment of Inertia:**\n - *Scenario:* A thin hoop has mass m = 12.0\,\text{kg}$and radius$r = 3.00\,\text{m}.\n    I = m r^2 = (12.0\,\text{kg})(3.00\,\text{m})^2 = (12.0)(9.00) = 108\,\text{kg}\cdot\text{m}^2\n\n- **Problem 17: Ranking Four Discrete Systems by Moment of Inertia:**\n - *Scenario:* Four systems (A, B, C, D), each containing four identical masses m$spaced along a lightweight rod with equal separation$d:\n - **System A:** The axis passes lengthwise directly through all four masses (r_i = 0$for all masses). Thus,$I_A = 0.\n - **System C:** The axis passes through the center of mass (midway between the central masses). The masses are as close to the axis as possible, minimizing \sum m r_i^2.\n - **System D:** The axis passes between an outer mass and a central mass, shifted away from the center of mass.\n - **System B:** The axis passes through an outermost mass at the far end, placing the opposite masses at the greatest possible distances (d, 2d, 3d).\n\n\n\n - *Ordering by Increasing Moment of Inertia:*\n    I_A < I_C < I_D < I_B$$