Rotational Dynamics, Torque Equilibrium, and Massive Pulley Systems

Course Administration and Examination Guidelines

• Unit Test 2 Scope and Weighting:

  • The second unit test accounts for $10\%$ of the total course grade.
  • Scheduled date: Next Thursday.
  • Content coverage: Encompasses all instructional material covered up to and including the upcoming Tuesday class.
  • Subject focus: Primarily concentrated on dynamics, with foundational elements of kinematics integrated.

• Examination Preparation Protocol:

  • A practice problem set containing questions modeled after Unit Test 2 is distributed approximately one week in advance.
  • The practice set provides numerical final answers but omits fully worked solutions to encourage independent problem-solving.
  • Preparation conditions must simulate the actual exam: students must work independently using only the approved formula sheet, avoiding third-party assistance such as ChatGPT.

• Curricular Progression for Pulley Systems:

  • Historical treatment: Prior units modeled pulleys as massless and frictionless (idealized systems).
  • Current treatment: Pulleys possess significant mass, requiring the inclusion of rotational inertia in equations of motion.
  • Experimental application: The setup from Experiment 5 (a cart on a track connected across a pulley to a hanging mass) is revisited to evaluate the physical consequences of a massive pulley.
  • Upcoming topic: The classic Atwood machine system will be analyzed next week to incorporate the rotational inertia of massive pulleys.

Foundations of Center of Mass and Extended Body Modeling

• Physical Concept of Center of Mass:

  • The center of mass ($\text{CM}$) represents the unique point corresponding to the weighted average position of all mass elements comprising an extended object.
  • In extended body diagrams, the entire gravitational force ($\text{weight}$) acting on the object is applied precisely at this point.

• Discrete Element Simulation Analysis:

  • For a single uniform square element of material, the center of mass resides at the exact geometric center due to equal spatial mass contribution.
  • As uniform mass squares are stacked linearly downward, the center of mass shifts systematically along the vertical axis toward the added mass.
  • For a uniform rod, the center of mass coincides with its geometric midpoint.

• Spatial Location of the Center of Mass:

  • The center of mass does not need to be physically located within the material boundaries of the body.
  • In a symmetric, hollow structure such as a ladder without steps (two parallel vertical rails) or a completed ladder, the center of mass lies along the line of geometric symmetry, situated in the open air between the side rails.
  • The center of mass is a mathematical position governed by the spatial distribution of mass, not a physical requirement of material presence.

Torque Mechanics and Effectiveness in Rotational Motion

• Mathematical Definition of Torque:

  • Torque ($\tau$) quantifies the tendency of an applied force to induce rotational motion about a designated pivot point.
  • Magnitude expression:     $\tau = r F_{\perp} = r F \sin(\theta)$
  • Parameter definitions:
  • $r$: Lever arm distance measured from the fixed pivot point to the application point of the force.
  • $F$: Magnitude of the applied force.
  • $F_{\perp}$: Component of the applied force oriented strictly perpendicular to the radial position vector $\mathbf{r}$.
  • $\theta$: Angle between the radial vector $\mathbf{r}$ and the applied force vector $\mathbf{F}$.

• Door Rotation Comparative Case Study:

  • A hinged door with four distinct forces applied evaluates rotational effectiveness:
  • Force $F_1$: Applied at the maximum distance from the hinge ($r_{\max}$) perpendicular to the door ($\theta = 90^\circ$, $F_{\perp} = F_1$). Produces the greatest torque magnitude and represents the most effective force for rotating the door.
  • Force $F_2$: Directed along the radial line toward the hinge ($\theta = 180^\circ$). Because $\sin(180^\circ) = 0$, $F_{\perp} = 0$, yielding zero torque ($\tau_2 = 0$). It is the least effective force, causing structural stress on the hinge without rotation.
  • Force $F_3$: Applied at the outer edge ($r_{\max}$) but inclined at an acute angle to the door surface. The parallel component ($F_{\parallel}$) produces zero torque; only the perpendicular component ($F_{\perp} = F_3 \sin(\theta)$) generates torque, making it less effective than $F_1$.
  • Force $F_4$: Applied perpendicularly ($\theta = 90^\circ$, $F_{\perp} = F_4$), but positioned close to the hinge ($r$ is small). Despite full perpendicularity, the small lever arm results in low torque.
  • Governing optimization principle: Maximizing torque requires applying force as far from the pivot as possible and at an angle perpendicular ($90^\circ$) to the lever arm.

Static Equilibrium Analysis of a Uniform Ladder

• System Parameters and Geometry:

  • Ladder structure: Uniform rigid rod of length $L = 5\,\text{m}$ leaning against a smooth, frictionless vertical wall.
  • Incline geometry: The ladder forms an angle of $\theta = 53.1^\circ$ relative to the horizontal ground.
  • Ladder weight: $W_{\text{ladder}} = 180\,\text{N}$, acting at the geometric center of mass ($\frac{L}{2} = 2.5\,\text{m}$ from the base).
  • Climber (guy) parameters: Person weighing $W_{\text{guy}} = 800\,\text{N}$ standing one-third of the way up the ladder ($\frac{L}{3} = \frac{5}{3}\,\text{m}$ from the base).
  • Contact interfaces:
  • Ground: Rough surface exerting an upward normal force ($n_{\text{base}}$) and a horizontal static friction force ($f_s$) directed rightward to prevent outward slipping.
  • Wall: Frictionless interface exerting a purely horizontal normal contact force ($n_{\text{wall}}$) directed to the left.

• Translational Static Equilibrium:

  • The vector sum of all external forces must equal zero ($\sum \mathbf{F} = \mathbf{0}$).
  • Horizontal axis ($x$-axis):     $\sum F_x = f_s - n_{\text{wall}} = 0$$f_s = n_{\text{wall}}$
  • Vertical axis ($y$-axis):     $\sum F_y = n_{\text{base}} - W_{\text{guy}} - W_{\text{ladder}} = 0$$n_{\text{base}} = W_{\text{guy}} + W_{\text{ladder}} = 800\,\text{N} + 180\,\text{N} = 980\,\text{N}$

• Rotational Static Equilibrium:

  • The vector sum of all torques about any reference pivot must equal zero ($\sum \tau = 0$).
  • Selecting the base of the ladder as the pivot eliminates torques from $n_{\text{base}}$ and $f_s$ because their lever arms are zero ($r = 0$).
  • Torques about the base:     $\tau_{\text{net}} = \tau_{\text{base}} + \tau_{\text{friction}} + \tau_{\text{guy}} + \tau_{\text{ladder}} + \tau_{\text{wall}} = 0$
  • Directional sign conventions:
  • Clockwise (negative): Weight forces ($W_{\text{guy}}$ and $W_{\text{ladder}}$).
  • Counterclockwise (positive): Wall normal force ($n_{\text{wall}}$).
  • Component torque formulation:     $- W_{\text{guy}} \left(\frac{L}{3}\right) \cos(53.1^\circ) - W_{\text{ladder}} \left(\frac{L}{2}\right) \cos(53.1^\circ) + n_{\text{wall}} L \sin(53.1^\circ) = 0$
  • Solving for the wall reaction force:     $n_{\text{wall}} = 268\,\text{N}$
  • From horizontal equilibrium:     $f_s = n_{\text{wall}} = 268\,\text{N}$

• Net Reaction Force at the Base:

  • Vector magnitude of the ground force ($F_{\text{base}}$) combining horizontal friction and vertical normal force:     $F_{\text{base}} = \sqrt{(f_s)^2 + (n_{\text{base}})^2} = \sqrt{(268\,\text{N})^2 + (980\,\text{N})^2} = 1016\,\text{N} = 1.01 \times 10^3\,\text{N}$
  • Directional angle ($\phi$) above the horizontal ground:     $\phi = \tan^{-1}\left(\frac{n_{\text{base}}}{f_s}\right) = \tan^{-1}\left(\frac{980\,\text{N}}{268\,\text{N}}\right) = 74.7^\circ$

• Coefficient of Static Friction:

  • To guarantee that the ladder does not slip, the available maximum static friction must satisfy $f_{s,\max} \ge f_s$:     $f_{s,\max} = \mu_s n_{\text{base}}$$\mu_s (980\,\text{N}) \ge 268\,\text{N}$$\mu_{s,\min} = \frac{268\,\text{N}}{980\,\text{N}} \approx 0.273$
  • If $\mu_s < 0.273$, the base slips outward; if $\mu_s > 0.273$, the contact possesses sufficient stickiness to maintain static equilibrium.

Rotational Dynamics and the Rotational Analog of Newton's Second Law

• Translational Versus Rotational Dynamic Analogs:

  • Translational formulation:     $a = \frac{F_{\text{net}}}{m}$
  • Acceleration is directly proportional to net force and inversely proportional to inertial mass ($m$).
  • Rotational formulation:     $\alpha = \frac{\tau_{\text{net}}}{I}$$\tau_{\text{net}} = I \alpha$
  • Angular acceleration ($\alpha$) is directly proportional to net external torque ($\tau_{\text{net}}$) and inversely proportional to moment of inertia ($I$).
  • Linearization: Plotting $\tau_{\text{net}}$ versus $\alpha$ yields a linear slope equal to the moment of inertia ($I$), directly mirroring the force versus linear acceleration slope ($m$).

• Physical Demonstration of Rotational Inertia:

  • Figure skater spin mechanics:
  • Winding up: Arms extended outward maximize mass distance from the rotational axis, creating a large moment of inertia ($I$).
  • Spin execution: Tucking arms tightly against the torso ("hugging oneself") relocates mass close to the central axis, drastically reducing $I$.
  • Effect: For a given angular momentum/torque input, reducing $I$ results in a dramatic increase in angular velocity and angular acceleration.
  • With arms outstretched, a spinning jump may yield only $268^\circ$ to $360^\circ$ of rotation before landing; pulling the limbs inward enables up to $7$ complete revolutions.
  • Initiation of rotation: An angular acceleration requires an external torque; when jumping, this external torque is exerted by the ground on the skater's feet.

Moment of Inertia and Rotational Inertia Dynamics

• Characteristics of Moment of Inertia ($I$):

  • Replaces inertial mass in rotational mechanics.
  • Dependent on total mass ($M$), physical dimensions (radius/length), geometric shape, and the distribution of mass relative to the axis of rotation.

• Moment of Inertia of a Solid Cylinder or Disk:

  • Formula for a uniform solid disk or thin circular cylinder rotating about its central symmetry axis:     $I = \frac{1}{2} M R^2$
  • Real pulleys are modeled as uniform disks using this relation.

• Dynamic Pulley Example Problem:

  • Pulley properties: Mass $M = 15\,\text{kg}$, treated as a disk.
  • Applied tangential force: A light cable unwinding without slipping or stretching is pulled with a force of $F = 9\,\text{N}$.
  • Net torque acting on the disk:     $\tau_{\text{net}} = 0.54\,\text{N}\cdot\text{m}$
  • Moment of inertia calculation:     $I = 0.5 \times (15\,\text{kg}) \times R^2$
  • Angular acceleration result:     $\alpha = \frac{\tau_{\text{net}}}{I} = 6\,\text{rad/s}^2$
  • Angular acceleration rate: The disk gains $6\,\text{rad/s}$ of angular velocity every second.
  • Tangential acceleration ($a_t$) at the disk rim:     $a_t = R \alpha = R (6\,\text{rad/s}^2)$

Torque Calculations on a Rotating Wheel

• Multi-Force System on a Pivot Wheel:

  • A wheel of radius $R = 0.35\,\text{m}$ is subjected to three distinct applied forces:
  • Force 1 ($11.9\,\text{N}$): Directed radially straight downward toward the central pivot.
    • Because the force vector aligns radially ($\theta = 0^\circ$ or $180^\circ$), $F_{\perp} = 0$.
    • Generated torque: $\tau_{11.9} = 0$.
  • Force 2 ($14.6\,\text{N}$): Applied at the rim ($r = 0.35\,\text{m}$) at an angle of $40^\circ$ relative to the position vector $\mathbf{r}$.
    • Radial component: $14.6 \cos(40^\circ)\,\text{N}$ produces zero torque.
    • Tangential component: $14.6 \sin(40^\circ)\,\text{N}$.
    • Rotation direction: Clockwise (conventionally negative torque).
    • Torque calculation: $\tau_{14.6} = - (0.35\,\text{m}) (14.6\,\text{N}) \sin(40^\circ)$.
  • Force 3 ($8.5\,\text{N}$): Applied at the outer edge perpendicular to the radius vector.
    • Generates counterclockwise (positive) rotation.
    • Torque calculation: $\tau_{8.5} = + (0.35\,\text{m}) (8.5\,\text{N})$.

• Line of Action and Motion Classification:

  • Force line of action passing through the center of mass:
  • Yields zero torque about the center of mass.
  • Induces pure translational motion; rotation is impossible.
  • Force line of action not passing through the center of mass:
  • Generates a non-zero torque about the center of mass.
  • Causes rotational acceleration.
  • Edge application with fixed pivot: If the object is pinned at the center and a force is applied tangentially at the perimeter, only pure rotational motion occurs.
  • Unconstrained body: If an unpinned object experiences an off-center line of action, a coupled hybrid motion (simultaneous translation and rotation) occurs.

Non-Ideal Massive Pulleys in Coupled Dynamic Systems

• Physical Contrast with Ideal Pulley Systems:

  • In ideal, massless pulleys, the string tension is uniform across both sides ($T_1 = T_2$).
  • In real, massive pulleys, generating angular acceleration ($\alpha$) demands a non-zero net torque ($\tau_{\text{net}} \ne 0$).
  • Consequently, the tension forces on either side of the pulley must be unequal ($T_2 > T_1$ for clockwise angular acceleration).

• Modified Newton's Second Law System Setup:

  • Object 1: Box of mass $m_1 = 12\,\text{kg}$ situated on a horizontal frictionless surface.
  • Connecting element: Light, unstretchable wire passing over a massive, frictionless pulley of radius $R$.
  • Object 2: Hanging mass suspended vertically from the other side of the pulley.
  • System motion upon release: The hanging weight accelerates downward, the horizontal box accelerates rightward with linear acceleration $a$, and the pulley undergoes clockwise angular acceleration $\alpha$.

• Free Body Diagram Formulations:

  • Horizontal Box ($12\,\text{kg}$):
  • Vertical equilibrium: Normal force ($n$) directed upward balances gravitational weight ($m_1 g$) downward ($n - m_1 g = 0$).
  • Horizontal motion: Single tension $T_1$ acts to the right ($T_1 = m_1 a$).
  • Hanging Mass:
  • Downward gravitational force ($m_2 g$) exceeds upward string tension $T_2$ ($m_2 g - T_2 = m_2 a$).
  • Massive Pulley (Extended Body Diagram):
  • Pulley gravitational weight ($W_p$) acts downward at the center of mass.
  • String tension $T_1$ acts tangentially to the left at the top edge.
  • String tension $T_2$ acts tangentially downward at the right outer edge.
  • Axle force ($F_{\text{axle}}$) acts upward and to the right on the center to maintain static translational equilibrium.

• Dynamic Equations for the Massive Pulley:

  • Translational Static Equilibrium:
  • The pulley rotates but does not undergo translational displacement:       $\sum F_x = F_{\text{axle}, x} - T_1 = 0 \implies F_{\text{axle}, x} = T_1$$\sum F_y = F_{\text{axle}, y} - W_p - T_2 = 0 \implies F_{\text{axle}, y} = W_p + T_2$
  • Rotational Dynamics:
  • The weight of the pulley ($W_p$) and the axle force ($F_{\text{axle}}$) act at the center of mass ($r = 0$), producing zero torque.
  • Only the unequal string tensions provide torque:       $\tau_{\text{net}} = T_2 R - T_1 R = I \alpha$
  • Kinematic Constraint:
  • For an unstretchable string that does not slip along the pulley groove, the tangential acceleration at the pulley rim equals the linear acceleration of the connected masses:       $\alpha = \frac{a_t}{R} = \frac{a}{R}$
  • Substitution into the rotational dynamic equation:       $(T_2 - T_1) R = I \left(\frac{a}{R}\right)$$T_2 - T_1 = I \frac{a}{R^2}$

Questions and Discussion

• Angle Representation in Wall Torque Calculation:

  • Question: In the ladder equilibrium equation, why is the torque due to the wall expressed with $\sin(53.1^\circ)$?
  • Response: The angle $53.1^\circ$ is given between the ladder and the horizontal floor. The horizontal normal force of the wall ($n_{\text{wall}}$) and the ladder create alternate interior angles when extending geometric lines. To compute torque, one needs the component of $n_{\text{wall}}$ perpendicular to the ladder, which corresponds to the opposite side of the right triangle ($n_{\text{wall}} \sin(53.1^\circ)$). The parallel component, $n_{\text{wall}} \cos(53.1^\circ)$, forms an angle of $180^\circ$ with the position vector and produces zero torque.

• Center of Mass versus Pivot Point for Torque:

  • Question: Does a force produce zero torque only when its line of action passes through the pivot point, or can it pass through the center of mass and not the pivot point?
  • Response: Torque is strictly defined relative to the chosen pivot point. A force produces zero torque if and only if its line of action passes through that specific pivot point. If a force passes through the center of mass but misses the selected pivot point, it produces a non-zero torque about that pivot.

• Impact of Pulley Weight on Torque:

  • Question: Does the weight of the pulley have no effect on torque since it pulls straight down through the center of mass?
  • Response: The weight of the pulley pulls directly through the central axis (the center of mass and pivot), meaning its lever arm is zero ($r = 0$), so it produces zero torque and cannot cause rotational acceleration. However, it affects translational equilibrium; without an upward supporting force from the axle, the pulley would accelerate downward in translation.

• Axle Reaction Force Nomenclature:

  • Question: Does the supporting force acting on the pulley's axle count as a normal force?
  • Response: It can be categorized conceptually as a normal contact force exerted by the axle mounting, but it is explicitly referred to as the axle force ($F_{\text{axle}}$), resolved into horizontal and vertical components that prevent translational acceleration.