Chapter 8 Lectures: Potential Energy and Conservation of Energy (Slides 25-34)

Principles of Mechanical Energy Conservation

• Definition of Total Mechanical Energy:

  • Mechanical energy ($E$) represents the sum of the kinetic energy ($K$) and total potential energy ($U$) present within a defined system:         $E = K + U$

  • Potential energy functions incorporated into mechanical systems typically include gravitational potential energy ($U_g$) and elastic or spring potential energy ($U_s$):         $U = U_g + U_s$

• Criteria for Conservation:

  • Mechanical energy is strictly conserved if and only if no net work is performed by non-conservative (dissipative) forces on the system:         $W_{nc} = 0$

  • When all non-conservative forces (such as friction, air drag, or non-elastic deformation) do zero work, the total mechanical energy remains constant throughout the motion of the system.

  • Total energy in the broadest thermodynamic sense accounts for thermal, chemical, and dissipative non-conservative energies, whereas mechanical energy accounts exclusively for macroscopic kinetic and potential energies.

• Mathematical Formulations:

  • Equating the initial mechanical energy state ($E_0$) to the final mechanical energy state ($E_f$):         $E_0 = E_f$

  • Expanding into constituent kinetic and potential energy components:         $K_0 + U_0 = K_f + U_f$

  • Reformulating in terms of discrete changes ($\Delta$) across states:         $0 = (K_f - K_0) + (U_f - U_0)$         $\Delta K + \Delta U = 0$

• Systematic Problem-Solving Framework (Energy Accounting Table):

  • Step 1: Clearly isolate and define the physical system.

  • Step 2: Verify that no non-conservative forces perform work ($W_{nc} = 0$).

  • Step 3: Construct an accounting matrix recording initial and final mechanical energy terms across each energy modality:

    • Row 1: Initial State ($0$)

    • Row 2: Final State ($f$)

    • Column 1: Kinetic Energy ($K = \frac{1}{2}mv^2$)

    • Column 2: Gravitational Potential Energy ($U_g = mgy$)

    • Column 3: Elastic Potential Energy ($U_s = \frac{1}{2}kx^2$)

  • Step 4: Set the sum of all initial energy columns equal to the sum of all final energy columns and isolate the desired unknown kinematic or dynamic variable.

Free Fall and Projectile Motion Analysis

• Problem Specification:

  • A rock is propelled from the edge of a cliff at an initial elevation of $y_0 = 140\,\text{m}$ above ground level.

  • The rock possesses an initial speed of $v_0 = 25\,\text{m/s}$.

  • The directional orientation (angle of inclination) of the initial velocity vector is unspecified.

  • Air resistance is assumed to be negligible ($W_{nc} = 0$).

  • The objective is to compute the final speed ($v_f$) of the rock evaluated immediately prior to impact with the ground ($y_f = 0\,\text{m}$).

• Boundary Conditions and Procedural Safeguards:

  • The ground elevation is designated as the origin ($y_f = 0\,\text{m}$).

  • The evaluation must occur strictly at the instant just before contact with the surface. Setting the speed at impact equal to zero represents a fundamental operational error, as coming to rest occurs after the collision process via dissipative non-conservative forces.

• Energy Accounting and Derivation:

  • Initial energy configuration:         $K_0 = \frac{1}{2}mv_0^2$         $U_{g,0} = mgy_0$

  • Final energy configuration:         $K_f = \frac{1}{2}mv_f^2$         $U_{g,f} = mgy_f = mg(0) = 0$

  • Balancing total mechanical energy:         $\frac{1}{2}mv_0^2 + mgy_0 = \frac{1}{2}mv_f^2$

  • Because the mass ($m$) appears linearly in every term, it cancels completely from the equation:         $\frac{1}{2}v_0^2 + gy_0 = \frac{1}{2}v_f^2$

  • Isolating final velocity ($v_f$):         $v_f^2 = v_0^2 + 2gy_0$         $v_f = \sqrt{v_0^2 + 2gy_0}$

• Quantitative Computation:

  • Substituting known numerical values using standard gravitational acceleration ($g = 9.8\,\text{m/s}^2$):         $v_f = \sqrt{(25\,\text{m/s})^2 + 2(9.8\,\text{m/s}^2)(140\,\text{m})}$         $v_f = \sqrt{625\,\text{m}^2/\text{s}^2 + 2744\,\text{m}^2/\text{s}^2} = \sqrt{3369\,\text{m}^2/\text{s}^2} \approx 58.04\,\text{m/s}$

  • Rounding to two significant figures yields:         $v_f = 58\,\text{m/s}$

• Path Independence of Conservative Forces:

  • The force of gravity ($F_g$) is a conservative vector field, meaning that the total work performed between two endpoints is strictly independent of the geometric path taken.

  • Whether the projectile is launched horizontally, inclined upward, directed downward, or forced through a complex curvilinear trajectory, the magnitude of the final impact speed is identical because the initial and terminal boundary heights ($y_0$ and $y_f$) govern the potential energy transformation.

Pendulum Dynamics and Height Geometry

• System Configuration and Physical Parameters:

  • A simple pendulum of cord length $L = 35.0\,\text{cm} = 0.350\,\text{m}$ is suspended from a fixed pivot.

  • The pendulum bob is elevated to an initial angular displacement of $\theta_0 = 45.0^\circ$ relative to the vertical line of suspension.

  • The bob is released from rest, setting initial velocity to zero:         $v_0 = 0\,\text{m/s}$

  • The target parameter is the linear tangential speed ($v_f$) when the pendulum swings through an intermediate angle of $\theta_f = 30.0^\circ$ from the vertical.

  • Air resistance and cord mass are neglected.

• Geometric Projection of Vertical Height:

  • The vertical coordinate system is established by defining the origin ($y = 0\,\text{m}$) at the lowest point of the swing arc (the equilibrium position of the point mass).

  • The vertical distance from the fixed pivot down to the bob at any angular displacement $\theta$ corresponds to the adjacent side of a right triangle of hypotenuse $L$:         $y_{\text{pivot-to-bob}} = L\cos(\theta)$

  • Because the total vertical distance from the pivot to the origin is $L$, the height ($y$) of the bob above the origin is obtained by subtracting the adjacent component from $L$:         $y = L - L\cos(\theta) = L(1 - \cos(\theta))$

  • Applying this relationship to both angular configurations:         $y_0 = L - L\cos(\theta_0)$         $y_f = L - L\cos(\theta_f)$

• Energy Conservation Formulation:

  • Initial state ($v_0 = 0\,\text{m/s}$):         $K_0 = 0$         $U_{g,0} = mgy_0 = mg(L - L\cos(\theta_0))$

  • Final state at $\theta_f$:         $K_f = \frac{1}{2}mv_f^2$         $U_{g,f} = mgy_f = mg(L - L\cos(\theta_f))$

  • Equating mechanical energy expressions:         $mg(L - L\cos(\theta_0)) = \frac{1}{2}mv_f^2 + mg(L - L\cos(\theta_f))$

  • Eliminating mass ($m$) from all terms:         $gL - gL\cos(\theta_0) = \frac{1}{2}v_f^2 + gL - gL\cos(\theta_f)$

  • Canceling the constant term $gL$ on both sides:         $-gL\cos(\theta_0) = \frac{1}{2}v_f^2 - gL\cos(\theta_f)$

  • Rearranging to isolate the final velocity $v_f$:         $\frac{1}{2}v_f^2 = gL(\cos(\theta_f) - \cos(\theta_0))$         $v_f = \sqrt{2gL(\cos(\theta_f) - \cos(\theta_0))}$

• Numerical Evaluation:

  • Gravitational acceleration is set to $g = 9.80\,\text{m/s}^2$.

  • Length of suspension is $L = 0.350\,\text{m}$.

  • Trigonometric parameters:         $\cos(30.0^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$         $\cos(45.0^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071$         $\Delta \cos(\theta) = 0.8660 - 0.7071 = 0.1589$

  • Computing linear speed:         $v_f = \sqrt{2(9.80\,\text{m/s}^2)(0.350\,\text{m})(0.8660 - 0.7071)}$         $v_f = 1.04\,\text{m/s}$

The Heavy Pendulum Demonstration: Physical Principles and Safety Implications

• Experimental Configuration:

  • A massive, solid sphere (such as a bowling ball) is suspended from an overhead ceiling mount by a steel cable or chain.

  • The demonstrator (e.g., astrophysicist Neil deGrasse Tyson) pulls the ball back until it directly contacts the tip of their nose, stands completely motionless against a wall, and releases the ball from rest ($v_0 = 0\,\text{m/s}$) without imparting an initial velocity.

• Theoretical Basis for Confidence:

  • Under the governing condition $W_{nc} \le 0$, total mechanical energy cannot increase beyond its initial value:         $E_f \le E_0$

  • Because the release speed is zero, the total initial energy consists exclusively of gravitational potential energy:         $E_0 = mgy_{\text{nose}}$

  • Upon executing a full oscillation and returning to the reversal point where $v = 0\,\text{m/s}$, all remaining mechanical energy converts back into potential energy:         $E_f = mgy_{\text{return}}$

  • In an ideal frictionless system, $y_{\text{return}} = y_{\text{nose}}$; in any real atmospheric environment where dissipative drag performs negative work ($W_{\text{drag}} < 0$), the return height is strictly less than the release height ($y_{\text{return}} < y_{\text{nose}}$).

  • The pendulum cannot physically reach or penetrate past the spatial coordinates of the demonstrator's nose, ensuring complete physical safety provided the demonstrator does not lean forward.

• Failure Modes and Perturbation Analysis:

  • The Hazard of an Initial Push: If the demonstrator imparts an initial velocity ($v_0 > 0\,\text{m/s}$) via an outward push, initial mechanical energy includes a kinetic term:         $E_0 = mgy_{\text{nose}} + \frac{1}{2}mv_0^2$

  • Upon the return swing, the point of zero kinetic energy occurs at an elevation exceeding the release elevation:         $y_{\text{return}} = y_{\text{nose}} + \frac{v_0^2}{2g}$

  • This extra elevation drives the heavy mass several inches past the release point, resulting in severe cranial trauma.

  • The Effect of Pulling Back Further: Pulling the mass farther back increases $y_0$ and the release angle; the pendulum will swing outward and return to that new elevated reference point.

  • Demonstration Protocols in Classroom Instruction: When executing this demonstration with students using smaller bob assemblies, strict protocols prohibit the student from shaking, bumping, or displacing the suspension line, as any mechanical perturbation imparts exogenous energy capable of altering the return boundary.

Vertical Spring-Mass System Under Gravitational Acceleration

• Problem Statement and System Parameters:

  • A solid block of mass $m = 2.50\,\text{kg}$ is positioned above a vertically mounted linear coil spring.

  • The block is dropped from rest ($v_0 = 0\,\text{m/s}$) from an initial height of $h = 1.50\,\text{m}$ above the uncompressed, relaxed upper surface of the spring.

  • The force constant of the spring is $k = 474\,\text{N/m}$.

  • The objective is to compute the maximum compression distance of the spring ($y_f$) occurring at the exact turning point where the descending block is brought momentarily to a complete stop ($v_f = 0\,\text{m/s}$).

• Coordinate System and Origin Alignment:

  • A single shared vertical axis is defined where the uncompressed equilibrium position of the top of the spring serves as the origin:         $y = 0\,\text{m}$         $x = 0\,\text{m}$

  • Upward displacement is defined as positive; downward displacement into the spring is negative.

  • Initial vertical position of the block:         $y_0 = +1.50\,\text{m}$

  • Initial spring strain displacement:         $x_0 = 0\,\text{m}$

  • At the bottom turning point, the spring is compressed by an amount equal to the negative vertical coordinate of the block:         $x_f = y_f$

• Energy Balance Formulation:

  • Initial state ($v_0 = 0\,\text{m/s}$, $x_0 = 0\,\text{m}$):         $K_0 = 0$         $U_{s,0} = 0$         $U_{g,0} = mgy_0$         $E_0 = mgy_0$

  • Final state at maximum compression ($v_f = 0\,\text{m/s}$):         $K_f = 0$         $U_{g,f} = mgy_f$         $U_{s,f} = \frac{1}{2}kx_f^2 = \frac{1}{2}ky_f^2$         $E_f = mgy_f + \frac{1}{2}ky_f^2$

  • Equating initial and final mechanical energy:         $mgy_0 = mgy_f + \frac{1}{2}ky_f^2$

• Quadratic Form Derivation and Dimensional Verification:

  • Rearranging terms into the standard quadratic form ($ay^2 + by + c = 0$):         $\frac{1}{2}ky_f^2 + mgy_f - mgy_0 = 0$

  • Evaluating numerical coefficients:         $a = \frac{1}{2}k = \frac{1}{2}(474\,\text{N/m}) = 237\,\text{N/m}$         $b = mg = (2.50\,\text{kg})(9.80\,\text{m/s}^2) = 24.5\,\text{N}$         $c = -mgy_0 = -(2.50\,\text{kg})(9.80\,\text{m/s}^2)(1.50\,\text{m}) = -36.75\,\text{N}\cdot\text{m}$

  • Assembled quadratic equation:         $(237\,\text{N/m})y_f^2 + (24.5\,\text{N})y_f - 36.75\,\text{N}\cdot\text{m} = 0$

  • Dimensional Consistency Check:

    • First term: $(\text{N/m}) \times (\text{m}^2) = \text{N}\cdot\text{m} = \text{J}$

    • Second term: $(\text{N}) \times (\text{m}) = \text{N}\cdot\text{m} = \text{J}$

    • Third term: $\text{N}\cdot\text{m} = \text{J}$

    • All terms are dimensionally homogeneous in units of mechanical energy (Joules).

• Analytical Solution and Root Extraction:

  • Applying the quadratic formula:         $y_f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$         $y_f = \frac{-24.5 \pm \sqrt{(24.5)^2 - 4(237)(-36.75)}}{2(237)}$         $y_f = \frac{-24.5 \pm \sqrt{600.25 + 34839}}{474}$         $y_f = \frac{-24.5 \pm \sqrt{35439.25}}{474} = \frac{-24.5 \pm 188.253}{474}$

  • Computing the two mathematical roots:         $y_{f,1} = \frac{-24.5 + 188.253}{474} = +0.34547\,\text{m}$         $y_{f,2} = \frac{-24.5 - 188.253}{474} = -0.448846\,\text{m}$

  • Physical Root Selection:

    • The positive root ($+0.345\,\text{m}$) corresponds to an unphysical extension of the free spring above equilibrium where tension would occur if the block were permanently attached.

    • Because the block compresses the spring downward beneath the origin, the physically meaningful coordinate is the negative value.

    • Rounding to three significant figures gives:             $y_f = -0.449\,\text{m} = -44.9\,\text{cm}$

    • The maximum compression distance of the spring from equilibrium is $44.9\,\text{cm}$.

Amusement Park Loop-the-Loop Mechanics

• Problem Specification:

  • A roller-coaster cart of mass $m = 1800\,\text{kg}$ travels along a frictionless track.

  • The cart is initially launched horizontally by releasing it from a compressed linear spring with spring constant $k = 1.5 \times 10^5\,\text{N/m}$.

  • The cart subsequently enters a vertical circular loop-the-loop of radius $R = 12\,\text{m}$.

  • The objective is to find the minimum distance ($x_0$) the spring must be compressed from its equilibrium position to ensure the cart completes the loop without detaching from the track.

• Dynamic Boundary Condition at the Apex (Loop Top):

  • The track is aligned vertically such that the top of the loop occurs at elevation:         $y_{\text{top}} = 2R = 2(12\,\text{m}) = 24\,\text{m}$

  • For the cart to maintain continuous contact with the track, the normal contact force ($F_N$) exerted downward by the track on the cart must be non-negative:         $F_N \ge 0$

  • Applying Newton's Second Law along the radial direction at the apex, taking downward toward the center of curvature as positive:         $\sum F_r = mg + F_N = m a_c = m\frac{v_f^2}{R}$

  • The limiting critical threshold for successfully negotiating the apex occurs when the cart is on the verge of separation, defined by the condition where the normal force approaches zero:         $F_N \rightarrow 0$

  • Setting $F_N = 0$ yields the minimal critical velocity required at the top:         $mg = m\frac{v_f^2}{R}$         $v_f^2 = gR$         $v_f = \sqrt{gR}$

• Conservation of Energy Formulations:

  • Initial State (Full Spring Compression):

    • The cart is held at rest: $v_0 = 0\,\text{m/s}$, so $K_0 = 0$

    • The base track elevation is set as the origin: $y_0 = 0\,\text{m}$, so $U_{g,0} = 0$

    • The spring is compressed by linear displacement $x_0$:             $E_0 = \frac{1}{2}kx_0^2$

  • Final State (Apex of the Loop):

    • The cart possesses minimal tangential speed: $v_f^2 = gR$             $K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}m(gR)$

    • The cart is at maximum elevation: $y_f = 2R$             $U_{g,f} = mgy_f = mg(2R) = 2mgR$

    • The cart has separated from the spring at its equilibrium position ($x_f = 0\,\text{m}$), meaning the spring retains zero stored elastic energy ($U_{s,f} = 0$).

    • Summing final mechanical energy terms:             $E_f = \frac{1}{2}mgR + 2mgR = \frac{5}{2}mgR$

• Algebraic Derivation of Minimum Compression:

  • Equating initial and final mechanical energy expressions:         $\frac{1}{2}kx_0^2 = \frac{5}{2}mgR$

  • Canceling the factor of $\frac{1}{2}$ and isolating $x_0$:         $kx_0^2 = 5mgR$         $x_0^2 = \frac{5mgR}{k}$         $x_0 = \sqrt{\frac{5mgR}{k}}$

• Numerical Calculation:

  • Inserting parameters ($m = 1800\,\text{kg}$, $g = 9.80\,\text{m/s}^2$, $R = 12\,\text{m}$, $k = 1.5 \times 10^5\,\text{N/m}$):         $x_0 = \sqrt{\frac{5(1800\,\text{kg})(9.80\,\text{m/s}^2)(12\,\text{m})}{1.5 \times 10^5\,\text{N/m}}}$         $x_0 = \sqrt{\frac{1058400}{150000}} = \sqrt{7.056} \approx 2.656\,\text{m}$

  • Rounding to two significant figures yields:         $x_0 = 2.7\,\text{m}$

Non-Conservative Dissipative Systems: Air Resistance and Terminal Velocity

• Case Study: Baseball Dropped from the Sears Tower:

  • A Chicago Cubs fan drops an official baseball of mass $m = 143\,\text{g} = 0.143\,\text{kg}$ from the observation roof of the Sears Tower.

  • The drop elevation above street level is $h = 443\,\text{m}$.

  • The ball is released from rest ($v_0 = 0\,\text{m/s}$).

  • Atmospheric drag is non-negligible, causing the baseball to reach a measured terminal speed of $v_t = 42.0\,\text{m/s}$.

  • The calculation targets the total mechanical energy dissipated ($E_{\text{lost}}$) by air resistance during descent.

• Mechanics of Terminal Velocity:

  • Aerodynamic drag ($F_{\text{drag}}$) is a non-conservative, reactionary force oriented antiparallel to the instantaneous velocity vector.

  • As falling velocity increases, the opposing air resistance magnitude grows until it precisely balances the gravitational weight of the object:         $\sum F_y = F_{\text{drag}} - mg = 0$         $F_{\text{drag}} = mg$

  • At this state of dynamic force equilibrium, the net acceleration of the object drops identically to zero ($a = 0\,\text{m/s}^2$).

  • Beyond this threshold, the object falls at a fixed, constant terminal velocity ($v_t$).

• Energy Balance and Dissipation Derivation:

  • Initial mechanical energy at the tower roof ($y_0 = 443\,\text{m}$, $v_0 = 0\,\text{m/s}$):         $E_0 = K_0 + U_{g,0} = 0 + mgy_0$         $E_0 = (0.143\,\text{kg})(9.80\,\text{m/s}^2)(443\,\text{m}) = 620.822\,\text{J}$

  • Kinetic energy attained at terminal velocity:         $K_{\text{terminal}} = \frac{1}{2}mv_t^2 = \frac{1}{2}(0.143\,\text{kg})(42.0\,\text{m/s})^2$         $K_{\text{terminal}} = 0.5 \times 0.143 \times 1764 = 126.126\,\text{J}$

  • Because the velocity remains clamped at $42.0\,\text{m/s}$ for the remaining descent, the kinetic energy cannot increase further. Upon reaching the ground ($y_f = 0\,\text{m}$), gravitational potential energy drops to zero:         $U_{g,f} = 0\,\text{J}$         $E_f = K_f = 126.126\,\text{J}$

  • Any gravitational potential energy that failed to convert into kinetic energy was dissipated into the atmosphere through the negative work of aerodynamic friction:         $E_{\text{lost}} = E_0 - E_f = 620.822\,\text{J} - 126.126\,\text{J} = 494.696\,\text{J}$

  • Rounding to three significant figures yields:         $E_{\text{lost}} = 495\,\text{J}$

Mathematical Derivation of the Generalized Work-Energy Theorem for an N-Particle System

• General Non-Conservative Energy Relation:

  • When non-conservative forces act within or upon a system, total mechanical energy is no longer conserved, and the net work done by non-conservative forces corresponds directly to the change in mechanical energy:         $W_{nc} = \Delta E = \Delta K + \Delta U$

  • For dissipative processes such as kinetic friction or drag, $W_{nc} < 0$, resulting in a continuous degradation of mechanical energy.

• Step-by-Step Derivation for an Arbitrary N-Particle System:

  • Step 1: Microscopic Force Balance on a Single Particle:

    • Consider a system composed of $N$ individual interacting point particles.

    • Isolating particle $i$, the net force ($\mathbf{F}_{\text{net}, i}$) acting upon it represents the vector sum of all internal forces exerted by the other $N-1$ particles plus all external forces.

    • By the fundamental Work-Energy Theorem, the net line integral of all forces acting across the infinitesimal trajectory $d\mathbf{r}_i$ of particle $i$ equals the change in that particle's kinetic energy:             $W_{\text{net}, i} = \int \mathbf{F}_{\text{net}, i} \cdot d\mathbf{r}_i = \Delta K_i$

  • Step 2: Summation Over All Particles in the System:

    • Summing scalar work contributions over all $N$ particles yields the total network done across the entire collection:             $\sum_{i=1}^N W_{\text{net}, i} = \sum_{i=1}^N \Delta K_i$

  • Step 3: Force Decomposition into Conservative and Non-Conservative Sectors:

    • The net work experienced by particle $i$ is split into conservative ($W_{c, i}$) and non-conservative ($W_{nc, i}$) components:             $W_{\text{net}, i} = W_{c, i} + W_{nc, i}$

    • Substituting this decomposition into the total system sum:             $\sum_{i=1}^N (W_{c, i} + W_{nc, i}) = \sum_{i=1}^N \Delta K_i$             $\sum_{i=1}^N W_{c, i} + \sum_{i=1}^N W_{nc, i} = \sum_{i=1}^N \Delta K_i$

  • Step 4: Formal Definition of the Potential Energy Function:

    • A potential energy function exists if and only if the force is conservative (the work around any closed path equals zero, $\oint \mathbf{F}_c \cdot d\mathbf{r} = 0$, proving path independence).

    • The change in potential energy ($\Delta U$) is defined mathematically as the negative of the work performed by that conservative force:             $W_{c, i} = -\Delta U_i$

  • Step 5: Substitution and Algebraic Rearrangement:

    • Replacing the conservative work terms with potential energy variations:             $-\sum_{i=1}^N \Delta U_i + \sum_{i=1}^N W_{nc, i} = \sum_{i=1}^N \Delta K_i$

    • Transposing the potential energy summation to the right side of the equality:             $\sum_{i=1}^N W_{nc, i} = \sum_{i=1}^N \Delta K_i + \sum_{i=1}^N \Delta U_i$

  • Step 6: Condensation into Macroscopic State Variables:

    • Defining total system work done by all non-conservative forces as $W_{nc} = \sum_{i=1}^N W_{nc, i}$, total system kinetic energy change as $\Delta K = \sum_{i=1}^N \Delta K_i$, and total system potential energy change as $\Delta U = \sum_{i=1}^N \Delta U_i$:             $W_{nc} = \Delta K + \Delta U$

    • If no non-conservative forces act ($W_{nc} = 0$), this rigorously re-establishes the conservation of mechanical energy:             $0 = \Delta K + \Delta U$

Questions & Discussion

• Interlocutor Question / Prompt:

  • "What you doing?"

• Speaker Response:

  • "I am doing my video."

• Interlocutor Follow-Up:

  • "Want me to let you go?"

• Speaker Response:

  • "It's okay. I'm gonna do homework anyways."

• Interlocutor Follow-Up:

  • "Oh, you are. You can't talk."

• Speaker Response:

  • "Thinking about them. Just… So one half m b not squared plus… I hear your homework. Oh, before we do that… Yeah. I can talk, but I'm just… We need to figure out… I don't know if want me to just let you go. So we need to figure out what the minimum speed is in order to make the loop."