Rotational Dynamics, Massive Pulleys, and Rolling Motion Without Slipping

Course Logistics and Academic Announcements

• Schedule and Laboratory Sessions:

  • There is no laboratory session scheduled for tomorrow.
  • Tomorrow is designated as a review problem-solving session to review problem sets prior to the test.
  • The assignment must be submitted tomorrow by 5:00 PM.

• Examination Details:

  • Unit Test 2 takes place on Thursday at 8:30 AM in Perthburg.
  • The first two items on the agenda are included in the scope of Unit Test 2.
  • Item number three on the agenda is grayed out; it will not be evaluated on Unit Test 2, but will appear on the final unit test.

• Personal Anecdotes and Principles:

  • Wordplay Anecdote: An observation regarding French vocabulary involving the word for cow ("vache") and horse ("cheval" / "shabal").
  • High School Reunion Reflections: A 20-year high school reunion highlights how quickly time passes as adulthood, careers, and family responsibilities take over, leading to lost contacts over two decades. Maintaining long-term friendships is an essential life practice.

Dynamics of Systems with Massive Pulleys

• Physical Distinctions of Massive Pulleys:

  • In introductory physics, pulleys are often modeled as massless and frictionless, meaning string tension is uniform throughout.
  • When a pulley has significant mass, it possesses rotational inertia and requires a net non-zero torque to undergo angular acceleration:         $\Sigma \tau \neq 0$

• If equal tensions act on opposite sides at identical radial distances and angles, the net torque is zero, preventing angular acceleration.

• To rotate, the tensions on either side of the pulley must be unequal:     $T_1 \neq T_2$

• Extended Free-Body Diagram for a Massive Pulley:

  • A massive pulley must be analyzed as an extended body rather than a point mass.
  • Tension Forces:
    • Tension $T_1$ acts tangentially at the rim at radius $R$ along the horizontal direction.
    • Tension $T_2$ acts tangentially at the rim at radius $R$ along the vertical direction.
  • Axle Support Force ($\mathbf{F}_{\text{axle}}$):
    • The axle force supports the pivot (center of mass) to balance the downward and horizontal loads.
    • Translational equilibrium along the horizontal axis ($x$-axis):             $F_{\text{axle}, x} = T_1$
    • Translational equilibrium along the vertical axis ($y$-axis):             $F_{\text{axle}, y} = T_2 + M_{\text{pulley}}g$
    • Because the line of action for both $\mathbf{F}_{\text{axle}}$ and the pulley's weight ($M_{\text{pulley}}g$) passes directly through the pivot point ($r = 0$), neither force exerts a torque about the center of mass.

• Mathematical Derivation for Box, Hanging Mass, and Massive Pulley System:

  • System Equations of Motion:
    • For the box of mass $M_{\text{box}}$ on a frictionless horizontal surface:             $F_N = M_{\text{box}}g$$T_1 = M_{\text{box}}a$
    • For the hanging mass $M_h$ with the downward vertical direction defined as positive:             $M_h g - T_2 = M_h a$$T_2 = M_h(g - a)$
    • For the rotating pulley (solid disk/cylinder) with clockwise rotation taken as positive:             $\Sigma \tau = I\alpha$$T_2 R - T_1 R = I\alpha$
  • Moment of Inertia and Kinematic Constraint:
    • Moment of inertia of a uniform solid disk/cylinder:             $I = \frac{1}{2} M_{\text{pulley}} R^2$
    • The relationship between linear tangential acceleration $a$ and angular acceleration $\alpha$:             $a = R\alpha$$\alpha = \frac{a}{R}$
  • Combining the Rotational Equations:
    • Substituting $I$ and $\alpha$ into the torque equation:             $(T_2 - T_1)R = \left(\frac{1}{2} M_{\text{pulley}} R^2\right)\left(\frac{a}{R}\right)$
    • Factoring and canceling $R$ on both sides yields:             $T_2 - T_1 = \frac{1}{2} M_{\text{pulley}} a$
  • Solving for Linear Acceleration ($a$):
    • Substitute the expressions for $T_1$ and $T_2$:             $M_h(g - a) - M_{\text{box}}a = \frac{1}{2} M_{\text{pulley}} a$
    • Rearranging terms:             $M_h g = \left(M_{\text{box}} + M_h + \frac{1}{2} M_{\text{pulley}}\right)a$
    • General form of linear acceleration for the system:             $a = \frac{M_h g}{M_{\text{box}} + M_h + \frac{1}{2} M_{\text{pulley}}}$
    • General System Law: Linear acceleration equals the net external pulling force divided by the total effective inertia of the system:             $a = \frac{F_{\text{pulling}}}{M_{\text{inertia, total}}}$

• Representative Numerical Calculations:

  • Linear acceleration of the system:         $a = 6.72\,\text{m/s}^2$
  • Tension $T_1$ acting on the box:         $T_1 = 32.7\,\text{N}$
  • Tension $T_2$ acting on the hanging mass:         $T_2 = 35.4\,\text{N}$
  • Difference between tensions:         $\Delta T = T_2 - T_1 = 35.4\,\text{N} - 32.7\,\text{N} = 2.7\,\text{N}$
  • This extra force imbalance of approximately $3\,\text{N}$ provides the net clockwise torque required to accelerate the massive pulley.

Analysis of the Modified Atwood Machine

• Physical Setup and Parameters:

  • Mass 1 (light mass): $m_1 = 2\,\text{kg}$
  • Mass 2 (heavy mass): $m_2 = 7\,\text{kg}$
  • Mass of the pulley: $M_{\text{pulley}} = 5\,\text{kg}$
  • Diameter of the pulley: $D = 15\,\text{cm} = 0.15\,\text{m}$
  • The string is massless and inextensible (does not stretch).
  • The pulley rotates without bearing friction.
  • Initial condition: Mass $m_1$ is given a sharp initial push vertically downward:         $v_0 = -2.40\,\text{m/s}\,\hat{j}$

• Dynamics and Free-Body Force Equations:

  • Because $m_2 > m_1$, the net driving gravitational torque accelerates the system toward $m_2$, making the acceleration vector directed upward for $m_1$ and downward for $m_2$.
  • Coordinate selection aligns the positive axis with the acceleration vector for each translating body.
  • Newton's Second Law for Mass 1 ($y$-axis pointing upward):$T_1 - m_1 g = m_1 a$$T_1 = m_1(g + a)$
  • Newton's Second Law for Mass 2 ($y$-axis pointing downward):$m_2 g - T_2 = m_2 a$$T_2 = m_2(g - a)$
  • Rotational Dynamic Equation for the Pulley:$(T_2 - T_1)R = I\alpha = \left(\frac{1}{2} M_{\text{pulley}} R^2\right)\left(\frac{a}{R}\right)$$T_2 - T_1 = \frac{1}{2} M_{\text{pulley}} a$

• Derivation and Calculation of Acceleration:

  • Substitute $T_1$ and $T_2$ into the simplified torque equation:         $m_2(g - a) - m_1(g + a) = \frac{1}{2} M_{\text{pulley}} a$$(m_2 - m_1)g = \left(m_1 + m_2 + \frac{1}{2} M_{\text{pulley}}\right)a$
  • Closed-form expression for acceleration:         $a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2} M_{\text{pulley}}}$
  • Numerical evaluation:         $m_2 - m_1 = 7\,\text{kg} - 2\,\text{kg} = 5\,\text{kg}$$m_{\text{total, effective}} = 2\,\text{kg} + 7\,\text{kg} + \frac{1}{2}(5\,\text{kg}) = 11.5\,\text{kg}$$a = \frac{5 \times 9.8}{11.5} = 4.26\,\text{m/s}^2$

• Kinematic Analysis of Mass 1:

  • Phase 1: Downward Deceleration to Turning Point:
    • Initial velocity is downward: $v_{0y} = -2.40\,\text{m/s}$.
    • Acceleration is upward: $a_y = +4.26\,\text{m/s}^2$.
    • Velocity and acceleration have opposite signs, causing the speed to decrease until $m_1$ instantaneously reaches its turning point where $v_y = 0$.
  • Maximum Downward Displacement (Distance Traveled):
    • Using the kinematic relation:             $v_y^2 = v_{0y}^2 + 2a_y \Delta y$$0 = (-2.40)^2 + 2(4.26)\Delta y$$\Delta y = -\frac{(-2.40)^2}{2(4.26)} = -0.676\,\text{m}$
    • The mass descends a total distance of $0.676\,\text{m}$ below its release point.
  • Velocity After an Elapsed Time of $t = 1.8\,\text{s}$:
    • Using the constant acceleration equation:             $v(t) = v_0 + at$$v(1.8) = -2.40 + (4.26)(1.8)$$v(1.8) = -2.40 + 7.668 = +5.27\,\text{m/s}$
    • The positive sign indicates that at $t = 1.8\,\text{s}$, mass 1 has reversed direction and is traveling vertically upward at a speed of $5.27\,\text{m/s}$.

Kinematics and Dynamics of Rolling Without Slipping

• Fundamental Definition of Rolling Without Slipping:

  • Rolling without slipping occurs when the instantaneous point of contact between a rolling circular object and the surface is at rest relative to the surface:         $v_{\text{contact}} = 0$
  • The contact point is held instantaneously stationary by the force of static friction ($f_s$).
  • Static friction exerts the torque necessary to rotate the object without slipping. Kinetic friction is absent in ideal rolling without slipping.

• Superposition of Motion (Translation and Rotation):

  • The complex motion of rolling without slipping decomposes into two distinct, independent components:
    1. Pure Translation of the Center of Mass: The center of mass moves linearly with velocity $v_{\text{cm}}$. All points on the object share this translational velocity.
    2. Pure Uniform Rotation about the Center of Mass: All points at distance $r$ rotate about the center of mass with angular velocity $\omega$, with tangential velocity given by $v_t = \omega r$.
  • Vector Summation across the Object:
    • At the Instantaneous Contact Point (Bottom):$v_{\text{bottom}} = v_{\text{cm}} - \omega R = 0$
    • At the Center of Mass:$v_{\text{center}} = v_{\text{cm}}$
    • At the Peak (Top of the Wheel):$v_{\text{top}} = v_{\text{cm}} + \omega R = 2 v_{\text{cm}}$

• Cycloid Trajectory:

  • The trajectory traced through space by a fixed point on the perimeter of a wheel rolling without slipping is a cycloid (a repeating succession of arches resembling half-sine waves).
  • At the apex of the cycloid, the velocity is purely horizontal and equal to $2 v_{\text{cm}}$.
  • At the cusp of the cycloid (ground contact), the velocity momentarily drops to zero as the point reverses vertical direction.

• Derivation of the Center-of-Mass Rolling Condition:

  • Over one complete rotational period ($T$), a wheel of radius $R$ rolls a linear distance equal to its circumference without sliding:         $\Delta x = 2\pi R$
  • Translational speed of the center of mass:         $v_{\text{cm}} = \frac{\Delta x}{\Delta t} = \frac{2\pi R}{T}$
  • By definition of angular frequency / angular velocity:         $\omega = \frac{2\pi}{T}$
  • Substituting $\omega$ into the linear velocity equation gives the rolling-without-slipping condition:         $v_{\text{cm}} = R\omega$
  • Differentiating with respect to time yields the corresponding acceleration condition:         $a_{\text{cm}} = R\alpha$

Rotational Inertia and the Rolling Incline Race

• Moments of Inertia for Standard Geometries of Identical Mass $M$ and Radius $R$:

  • Thin Spherical Shell / Hoop (Thin Ring):$I_{\text{hoop}} = 1.0\,M R^2$
    • All mass is concentrated at the outermost boundary radius $R$.
  • Solid Uniform Cylinder / Disk:$I_{\text{disk}} = \frac{1}{2} M R^2 = 0.5\,M R^2$
  • Uniform Solid Sphere:$I_{\text{sphere}} = \frac{2}{5} M R^2 = 0.4\,M R^2$

• Physical Significance of Mass Distribution:

  • Moment of inertia measures resistance to rotational acceleration.
  • As an object descends an incline, gravitational potential energy transforms into both translational kinetic energy ($\frac{1}{2} M v_{\text{cm}}^2$) and rotational kinetic energy ($\frac{1}{2} I \omega^2$).
  • Objects with higher moments of inertia require more torque and consume a larger fraction of available energy to achieve rotational speed, leaving less energy available for translational acceleration.

• The Incline Race Demonstration:

  • Two cylindrical objects of identical outer dimensions and mass ($M \approx 500\,\text{g}$, specifically $496\,\text{g}$) start simultaneously from rest at the top of an incline.
  • Object A has an internally weighted structure resulting in a low moment of inertia:         $I_A = 1.43 \times 10^{-3}\,\text{kg}\,\text{m}^2$
  • Object B has mass concentrated near the rim, resulting in a high moment of inertia:         $I_B = 2.78 \times 10^{-3}\,\text{kg}\,\text{m}^2$
  • Object A reaches the bottom of the incline first because its lower rotational inertia requires less torque to rotate, allowing a larger portion of energy to go toward linear acceleration.

Dynamic Derivation: Solid Sphere Rolling Down an Incline

• System Setup and Modeling:

  • A bowling ball is modeled as a uniform solid sphere of mass $M$, radius $R$, and moment of inertia:         $I = \frac{2}{5} M R^2$
  • The ball rolls without slipping down an incline angled at $\theta$ above the horizontal.
  • Coordinate system: The $x$-axis is tilted parallel to the ramp and points down the incline; the $y$-axis is perpendicular to the ramp pointing outward.

• Forces Acting on the Extended Free-Body Diagram:

  • Gravity ($Mg$): Acts directly downward through the center of mass.
  • Normal Force ($F_N$): Acts at the contact point, perpendicular to the incline, directed toward the center of mass.
  • Static Friction ($f_s$): Acts at the contact point, directed up the incline, providing the clockwise torque that causes rolling.

• Equations of Motion:

  • Newton's First Law Perpendicular to the Incline ($y$-axis):$F_N - Mg\cos(\theta) = 0$$F_N = Mg\cos(\theta)$
  • Newton's Second Law Parallel to the Incline ($x$-axis):$Mg\sin(\theta) - f_s = M a_{\text{cm}}$
  • Rotational Dynamics About the Center of Mass:
    • Gravity and the normal force produce zero torque about the center of mass because their lines of action pass through the pivot ($r = 0$ and $\sin(180^\circ) = 0$, respectively).
    • Static friction acts at distance $R$ with an angle of $90^\circ$ to the radial line:             $\Sigma \tau = f_s R = I\alpha$

• Coupling and Solution for Acceleration and Static Friction:

  • Applying the non-slip constraint $\alpha = \frac{a_{\text{cm}}}{R}$:         $f_s R = \left(\frac{2}{5} M R^2\right)\left(\frac{a_{\text{cm}}}{R}\right)$$f_s R = \frac{2}{5} M R a_{\text{cm}}$$f_s = \frac{2}{5} M a_{\text{cm}}$
  • Substitute $f_s$ back into the translational equation:         $Mg\sin(\theta) - \frac{2}{5} M a_{\text{cm}} = M a_{\text{cm}}$$Mg\sin(\theta) = \left(1 + \frac{2}{5}\right) M a_{\text{cm}}$$Mg\sin(\theta) = \frac{7}{5} M a_{\text{cm}}$
  • Solving for the center-of-mass linear acceleration:         $a_{\text{cm}} = \frac{5}{7} g\sin(\theta)$
  • Solving for the magnitude of static friction:         $f_s = \frac{2}{5} M \left(\frac{5}{7} g\sin(\theta)\right) = \frac{2}{7} Mg\sin(\theta)$

• Comparison to Frictionless Translation:

  • A frictionless point mass sliding down an inclined plane accelerates at:         $a_{\text{slide}} = g\sin(\theta)$
  • A rolling sphere accelerates at only $\frac{5}{7} \approx 0.714$ of the sliding value because $28.6\%$ of the available gravitational driving force is diverted into providing rotational torque.

Questions and Discussion

• Review Session and Test Clarification:

  • Question (Amna): Inquiries regarding the schedule for the following day and the content of the upcoming test.
  • Answer: The following day has no lab; instead, a problem review session will be held. The assignment is due at 5:00 PM. Unit Test 2 takes place on Thursday at 8:30 AM in Perthburg. Item 3 from the agenda is omitted from Test 2 and reserved for the final unit test.

• Radius Algebraic Cancellation:

  • Question: How did $R$ cancel out on both sides of the torque equation if one side had $R^2$ and the other had an $R$ in the denominator?
  • Answer: The torque equation contains $R$ on the left-hand side ($(T_2 - T_1)R$). On the right-hand side, $I\alpha = (\frac{1}{2} M R^2)(\frac{a}{R}) = \frac{1}{2} M R a$. Because both sides are multiplied by $R$ to the first power, dividing both sides by $R$ eliminates it entirely from the resulting linear equation.

• Signs in Free-Body Diagrams:

  • Question (Lucia): Why do tensions appear in opposite directions on the free-body diagrams, and should the tension on the pulley be negative?
  • Answer: Newton's Third Law dictates that the tension exerted by the string on the mass is directed opposite to the tension exerted by the string on the pulley. When analyzing each body separately, the signs in the equations of motion are determined strictly by the chosen coordinate frame for that specific isolated object.

• Validity of the Thin Hoop Inertia Formula:

  • Question (Darcy): Under what condition does an object's moment of inertia equal exactly $1.0\,M R^2$?
  • Answer: The expression $I = 1.0\,M R^2$ applies in the theoretical limit where the radial wall thickness of the ring or hoop is infinitesimally small, meaning the entirety of the object's mass is located at the outer radius $R$.

• Perpendicular Force Component in Tension:

  • Question: Is tension considered an exactly perpendicular force when calculating torque on the pulley?
  • Answer: Yes. Because the string departs tangentially from the circular rim of the pulley, the angle between the radial position vector from the pivot and the line of tension is always exactly $90^\circ$, making $\sin(90^\circ) = 1$.