PHYS 108A -Lecture 18: Torque, Rotational Kinematics, and Moment of Inertia

Static Equilibrium: Forces and Torques in Mechanical and Biomechanical Systems

• Equilibrium Conditions for Rigid Bodies:

  • A rigid body remains in static equilibrium if and only if both the net external force and the net external torque acting on the system equal zero simultaneously:     $\vec{F}_{\text{net}} = \sum \vec{F} = 0$     $\vec{\tau}_{\text{net}} = \sum \vec{\tau} = 0$

  • The force equilibrium condition guarantees that the center of mass has zero linear acceleration ($\vec{a} = 0$).

  • The torque equilibrium condition guarantees that the body has zero angular acceleration ($\vec{\alpha} = 0$) about any chosen pivot axis.

• Static Equilibrium Analysis of a Supported Beam:

  • System Geometry and Applied Loads:

  • A horizontal uniform beam is anchored to a vertical wall via a pivot pin at its left end ($x = 0$).

  • A supporting cable/rope is attached to the beam at a distance of $x_T = 0.500\,\text{m}$ from the pivot, extending upwards and to the left at an angle of $\theta = 80^\circ$ above the horizontal.

  • A load mass of $m = 2.00\,\text{kg}$ is placed at a distance of $x_W = 2.50\,\text{m}$ from the pivot pin.

  • The acceleration due to gravity is taken as $g = 9.8\,\text{m/s}^2$.

  • The downward gravitational force (weight) of the load is:       $W = mg = (2.00\,\text{kg})(9.8\,\text{m/s}^2) = 19.6\,\text{N}$

  • Tension Vector Decomposition:

  • The tension vector $\vec{T}$ acts at an angle of $80^\circ$ relative to the positive horizontal axis directed toward the wall:       $T_\perp = T_y = T\sin(80^\circ)$       $T_\parallel = T_x = -T\cos(80^\circ)$

  • Torque Balance About the Pivot Pin ($\sum \vec{\tau} = 0$):

  • Selecting the pivot pin at the wall ($x = 0$) eliminates the unknown normal reaction force $\vec{N}$ from the torque equation because its lever arm is zero ($\vec{\tau}_N = 0$):       $\vec{\tau}_{\text{net}} = \vec{\tau}_N + \vec{\tau}_T + \vec{\tau}_W = 0$       $\vec{\tau}_T + \vec{\tau}_W = 0$

  • Substituting the perpendicular lever arms and forces:       $(x_T) T\sin(80^\circ) - (x_W)(mg)\sin(90^\circ) = 0$       $(0.500\,\text{m}) T\sin(80^\circ) - (2.50\,\text{m})(2.00\,\text{kg})(9.8\,\text{m/s}^2)\sin(90^\circ) = 0$

  • Calculating the clockwise gravitational torque:       $\tau_W = (2.50\,\text{m})(19.6\,\text{N}) = 49.0\,\text{N}\cdot\text{m}$

  • Equating counterclockwise and clockwise torques:       $(0.500\,\text{m}) T\sin(80^\circ) = 49.0\,\text{N}\cdot\text{m}$       $T\sin(80^\circ) = \frac{49.0\,\text{N}\cdot\text{m}}{0.500\,\text{m}} = 98.0\,\text{N}$

  • Solving for the cable tension $T$:       $T_y = 98.0\,\text{N}$       $T = \frac{98.0\,\text{N}}{\sin(80^\circ)} = 99.5\,\text{N}$

  • Determining the horizontal component of tension $T_x$:       $T_x = -T\cos(80^\circ) = -(99.51\,\text{N})\cos(80^\circ) = -17.3\,\text{N}$

  • Force Balance and Pivot Reaction Force ($\sum \vec{F} = 0$):

  • Expressing translational equilibrium starting from Newton\'s second law ($\vec{F}_{\text{net}} = m\vec{a} = 0$):       $\vec{T} + \vec{N} + \vec{W} = 0$

  • Horizontal component balance:       $T_x + N_x + W_x = 0$       $-17.3\,\text{N} + N_x + 0 = 0$       $N_x = 17.3\,\text{N}$

  • Vertical component balance:       $T_y + N_y + W_y = 0$       $T_y + N_y - mg = 0$       $N_y = -T_y + mg$       $N_y = -98.0\,\text{N} + (2.00\,\text{kg})(9.8\,\text{m/s}^2)$       $N_y = -98.0\,\text{N} + 19.6\,\text{N} = -78.4\,\text{N}$

  • The wall pivot exerts a reaction force of $17.3\,\text{N}$ to the right (positive $x$) and $78.4\,\text{N}$ downward (negative $y$) to keep the beam in equilibrium.

• Biomechanical Analogy: The Human Forearm as a Third-Class Lever:

• The structural arrangement of the pinned beam directly mirrors the musculoskeletal mechanics of the human arm holding a weight:

  • The elbow joint acts as the pivot axis ($x = 0$).

  • The biceps brachii (lifting muscle) inserts via its tendon into the radius at a short distance ($4.0\,\text{cm}$) from the elbow joint, analogous to the cable attached at $0.500\,\text{m}$.

  • The external load is supported in the hand at a distance of $35\,\text{cm}$ from the elbow joint, analogous to the weight placed at $2.50\,\text{m}$.

• Mechanical Disadvantage and High Muscle Forces:

  • Because the lever arm of the muscle tendon ($4.0\,\text{cm}$) is substantially shorter than the load arm ($35\,\text{cm}$), the mechanical advantage is much less than 1:       $\text{Mechanical Advantage} = \frac{r_{\text{effort}}}{r_{\text{load}}} = \frac{4.0\,\text{cm}}{35\,\text{cm}} \approx 0.114$

  • Consequently, the internal tensile force exerted by the biceps muscle must be vastly larger than the external weight supported by the hand to establish torque equilibrium ($\tau_{\text{muscle}} = \tau_{\text{load}}$).

  • The downward joint reaction force at the elbow joint ($N_y$) must be correspondingly large to balance the large upward muscle pull, creating significant joint compression.

Rotational Kinematics and Angular Variables

• Fundamental Rotational Quantities:

  • Angular Position ($\theta$):

  • Measures rotational displacement relative to a reference axis.

  • Expressed in radians ($\text{rad}$), which is a dimensionless ratio of arc length to radius ($\theta = \frac{s}{r}$).

  • Angular Velocity ($\omega$):

  • Rate of change of angular position with respect to time:       $\omega = \frac{\Delta \theta}{\Delta t}$

  • Expressed in units of radians per second ($\text{rad/s}$).

  • Conversion factor for revolutions: $1\,\text{rev} = 2\pi\,\text{rad}$.

  • For a system rotating at a given frequency in revolutions per second:       $\omega = 2\pi \cdot \left(\frac{\text{rev}}{\text{s}}\right)$

  • Angular Acceleration ($\alpha$):

  • Rate of change of angular velocity with respect to time:       $\alpha = \frac{\Delta \omega}{\Delta t}$

  • Expressed in units of radians per second squared ($\text{rad/s}^2$).

• Relationships Between Linear and Angular Motion:

  • Points situated at a radial distance $r$ from the axis of rotation travel along circular paths.

  • Arc Length / Distance ($s$):     $s = r\theta$     $\Delta s = r\Delta \theta$

  • Tangential Velocity ($v$):

  • Differentiating arc length with respect to time yields linear speed along the tangential direction:       $v = \frac{\Delta s}{\Delta t} = r\left(\frac{\Delta \theta}{\Delta t}\right) = r\omega$

  • While all points on a rigid rotating body share the exact same angular velocity $\omega$, their linear speed $v$ scales linearly with radial distance $r$.

  • Tangential Acceleration ($a_t$):

  • Differentiating tangential velocity with respect to time produces tangential acceleration, representing changes in linear speed:       $a_t = \frac{\Delta v}{\Delta t} = r\left(\frac{\Delta \omega}{\Delta t}\right) = r\alpha$

  • Centripetal (Radial) Acceleration ($a_c$):

  • Arises from the continuous change in the direction of the velocity vector during circular motion:       $a_c = \frac{v^2}{r} = \frac{(r\omega)^2}{r} = \omega^2 r$

  • Directed radially inward toward the axis of rotation.

• Translational Versus Rotational Kinematics Analogy:

  • Constant acceleration kinematic equations map directly from translational coordinates to rotational coordinates:

  • Position / Displacement:

    • Translational: $x\,(\text{m})$

    • Rotational: $\theta\,(\text{rad})$

  • Velocity:

    • Translational: $v\,(\text{m/s})$

    • Rotational: $\omega\,(\text{rad/s})$

  • Acceleration:

    • Translational: $a\,(\text{m/s}^2)$

    • Rotational: $\alpha\,(\text{rad/s}^2)$

  • Standard Kinematic Formulations (Constant Acceleration):

  • Equation I (Displacement with Average Velocity):       $x(t) = x_0 + \bar{v}t$       $\theta(t) = \theta_0 + \bar{\omega}t$

  • Equation II (Average Velocity Definition):       $\bar{v} = \frac{v + v_0}{2}$       $\bar{\omega} = \frac{\omega + \omega_0}{2}$

  • Equation III (Velocity as a Function of Time):       $v = v_0 + at$       $\omega = \omega_0 + \alpha t$

  • Equation IV (Position as a Function of Time):       $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$       $\theta(t) = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$

  • Equation V (Velocity as a Function of Displacement):       $v^2 - v_0^2 = 2a(x - x_0)$       $\omega^2 - \omega_0^2 = 2\alpha(\theta - \theta_0)$

Dynamics of Rotational Motion and Newton\'s Second Law for Rotation

• Derivation of Torque for a Point Mass:

  • Consider a point mass $m$ constrained to move along a circular path of radius $r$.

  • If a tangential force $F_t$ is applied tangent to the circular trajectory, Newton\'s second law dictates:     $F_t = m a_t$

  • Substituting tangential acceleration in terms of angular acceleration ($a_t = r\alpha$):     $F_t = m (r\alpha)$

  • Multiplying both sides by the radial lever arm $r$ gives the net applied torque $\tau$:     $\tau = r F_t = r(mr\alpha) = mr^2\alpha$

  • The quantity $mr^2$ is the rotational inertia (moment of inertia $I$) of the point mass:     $\tau = I\alpha$

• Newton\'s Second Law for Rotation:

  • The rotational analogue of Newton\'s second law ($\vec{F}_{\text{net}} = m\vec{a}$) is:     $\vec{\tau}_{\text{net}} = I\vec{\alpha}$

  • Torque (expressed in $\text{N}\cdot\text{m}$) produces angular acceleration $\alpha$ ($\text{rad/s}^2$), with the moment of inertia $I$ ($\text{kg}\cdot\text{m}^2$) acting as the inertial resistance to changes in rotational state.

Moment of Inertia for Discrete Mass Systems

• Definition of Moment of Inertia for Discrete Mass Distributions:

  • For a system of $N$ point masses rigidly connected at distances $r_i$ from a specified axis of rotation:     $I = \sum_{i=1}^{N} m_i r_i^2$

  • Moment of inertia depends on:

  • The total mass of the system.

  • The geometric distribution of the mass relative to the chosen axis.

  • The location and orientation of the axis of rotation itself.

• Radial Scaling Analysis (Single Point Mass):

  • For a mass $m$ rotating at radius $r_A = 2r$ versus $r_B = r$ with identical angular acceleration $\alpha$:     $\tau_A = m(2r)^2\alpha = 4mr^2\alpha$     $\tau_B = mr^2\alpha$

  • Doubling the radial distance increases the required torque by a factor of 4 ($2^2 = 4$):     $\frac{\tau_A}{\tau_B} = 4$

• "Shish Kebab / Dango" Discrete System Analysis:

  • Consider an object consisting of three identical masses $m$ positioned along a light rigid rod.

  • Case 1: Pivot at One End (End Axis):

  • The three masses are located at distances $r_1 = 0$ (at the pivot), $r_2 = r$, and $r_3 = 2r$ from the rotation axis.

  • The total torque required to produce angular acceleration $\alpha$ is:       $\tau = m(0)^2\alpha + m(r)^2\alpha + m(2r)^2\alpha$       $\tau = 0 + mr^2\alpha + 4mr^2\alpha = 5mr^2\alpha$

  • Total moment of inertia about the end axis:       $I_{\text{end}} = 5mr^2$

  • Reformulation in terms of Total Mass ($M$) and Total Length ($\ell$):

    • Total mass: $M = 3m \implies m = \frac{M}{3}$.

    • Total length of the assembly: $\ell = 2r \implies r = \frac{\ell}{2}$.

    • Substituting into the torque expression:         $\tau = 5\left(\frac{M}{3}\right)\left(\frac{\ell}{2}\right)^2 \alpha = 5\left(\frac{M}{3}\right)\left(\frac{\ell^2}{4}\right)\alpha = \frac{5}{12}M\ell^2\alpha$         $I_{\text{end}} = \frac{5}{12}M\ell^2$

  • Case 2: Pivot at the Center Mass (Central Axis):

  • If the axis of rotation passes through the central mass, the mass positions are $r_1 = -r$, $r_2 = 0$, and $r_3 = +r$.

  • The total torque required is:       $\tau = m(-r)^2\alpha + m(0)^2\alpha + m(r)^2\alpha = 2mr^2\alpha$

  • Total moment of inertia about the central axis:       $I_{\text{center}} = 2mr^2$

  • Comparison Between End Axis and Central Axis:

  • Rotating the object about one end requires significantly more torque than rotating it about its center for the same angular acceleration:       $\frac{\tau_{\text{end}}}{\tau_{\text{center}}} = \frac{5mr^2\alpha}{2mr^2\alpha} = \frac{5}{2} = 2.5$

  • The end-pivoted configuration requires $2.5\times$ (or $150\%$ more) torque because a larger fraction of its mass is situated farther from the axis of rotation.

Rotational Inertia of Continuous Rigid Bodies

• Rotational Inertia of Standard Continuous Geometries:

• For continuous mass distributions of uniform density, the moment of inertia is computed by integrating over the body volume: $I = \int r^2\,dm$.

• Standard Moments of Inertia Reference Values:

  • Thin Hoop (or Thin Cylindrical Shell) about Cylinder Axis:       $I = MR^2$

  • Annular Cylinder (Ring) with Inner Radius $R_1$ and Outer Radius $R_2$ about Cylinder Axis:       $I = \frac{1}{2}M(R_1^2 + R_2^2)$

  • Solid Cylinder (or Uniform Disk) about Cylinder Axis:       $I = \frac{1}{2}MR^2 = 0.50\,MR^2$

  • Solid Cylinder (or Disk) about Central Diameter:       $I = \frac{1}{4}MR^2 + \frac{1}{12}M\ell^2$

  • Thin Rod of Length $\ell$ about Perpendicular Axis Through Center:       $I = \frac{1}{12}M\ell^2$

  • Thin Rod of Length $\ell$ about Perpendicular Axis Through One End:       $I = \frac{1}{3}M\ell^2$

  • Solid Sphere about Any Diameter:       $I = \frac{2}{5}MR^2 = 0.40\,MR^2$

  • Thin Spherical Shell about Any Diameter:       $I = \frac{2}{3}MR^2 \approx 0.67\,MR^2$

  • Thin Hoop about Any Diameter:       $I = \frac{1}{2}MR^2$

  • Rectangular Slab (Dimensions $a \times b$) about Perpendicular Axis Through Center:       $I = \frac{1}{12}M(a^2 + b^2)$

  • Comparative Resistance to Rotational Acceleration:

• For objects possessing the same total mass $M$ and outer radius $R$, the geometric coefficient $c$ in $I = cMR^2$ dictates resistance to angular acceleration:

  • Least Resistant: Solid Sphere ($c = \frac{2}{5} = 0.40$)

  • Second Least Resistant: Solid Disk / Cylinder ($c = \frac{1}{2} = 0.50$)

  • Second Most Resistant: Thin Spherical Shell ($c = \frac{2}{3} \approx 0.67$)

  • Most Resistant: Thin Hoop / Cylindrical Shell ($c = 1.00$)

• Mass concentrated closer to the axis of rotation results in a lower moment of inertia, requiring less torque for a given angular acceleration.

Rolling Motion and Energy Conservation

• Mechanics of Rolling Without Slipping:

• Role of Static Friction:

  • Static friction ($f_s$) between the rolling body and the surface is necessary to provide the torque that initiates and maintains rotational motion.

  • When rolling without slipping occurs, the instantaneous velocity of the contact point relative to the surface is identically zero ($v_{\text{contact}} = 0$).

  • Because the contact point does not slide, static friction does no mechanical work ($W = 0$) and does not dissipate mechanical energy into heat.

• No-Slip Condition:

  • The translational velocity of the center of mass $v$ is strictly coupled to the angular velocity $\omega$ by:       $v = r\omega \iff \omega = \frac{v}{r}$

  • If $\omega \neq \frac{v}{r}$, the contact point slips relative to the surface, and kinetic friction actively dissipates energy.

  • Total Mechanical Energy in Rolling Motion:

• The total kinetic energy of an object rolling without slipping combines translational kinetic energy of its center of mass and rotational kinetic energy about its center of mass:     $KE_{\text{total}} = KE_{\text{translational}} + KE_{\text{rotational}}$     $KE_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

• Substituting the rolling without slipping condition ($\omega = \frac{v}{r}$):     $KE_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2$

  • Conservation of Energy on an Incline of Height $h$:

• When an object of mass $m$ and radius $R$ is released from rest at height $h$ on an inclined ramp with friction:     $KE_i + PE_i = KE_f + PE_f$     $0 + mgh = \left(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\right) + 0$     $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2$

• Because mechanical energy is conserved, all objects with identical mass $m$ released from the same height $h$ acquire the exact same total kinetic energy at the base of the ramp:     $KE_{\text{total}, f} = mgh$

• However, the distribution of energy between translational and rotational forms depends entirely on the body\'s moment of inertia.

  • Analytical Derivations of Final Translational Velocity ($v$):

• Parameter baseline: $m = 1.0\,\text{kg}$, $R = 3.0 \times 10^{-3}\,\text{m}$.

• Derivation for a Solid Sphere ($I = \frac{2}{5}mr^2$):     $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$     $mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$     $mgh = \left(\frac{1}{2} + \frac{1}{5}\right)mv^2 = \left(\frac{5}{10} + \frac{2}{10}\right)mv^2 = \frac{7}{10}mv^2$     $gh = \frac{7}{10}v^2$     $v^2 = \frac{10}{7}gh$     $v = \sqrt{\frac{10}{7}gh} \approx 1.195\sqrt{gh}$

• Derivation for a Solid Disk / Cylinder ($I = \frac{1}{2}mr^2$):     $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$     $mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$     $mgh = \left(\frac{1}{2} + \frac{1}{4}\right)mv^2 = \left(\frac{2}{4} + \frac{1}{4}\right)mv^2 = \frac{3}{4}mv^2$     $gh = \frac{3}{4}v^2$     $v^2 = \frac{4}{3}gh$     $v = \sqrt{\frac{4}{3}gh} \approx 1.155\sqrt{gh}$

• Derivation for a Thin Spherical Shell ($I = \frac{2}{3}mr^2$):     $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$     $v = \sqrt{\frac{6}{5}gh} \approx 1.095\sqrt{gh}$

• Derivation for a Thin Hoop ($I = mr^2$):     $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$     $v = \sqrt{gh} = 1.000\sqrt{gh}$

  • Implications for the Race Down an Incline:

• Velocity at the bottom is independent of both total mass $m$ and radius $R$, depending solely on the shape factor $c$ in $I = cmr^2$:     $v = \sqrt{\frac{2gh}{1 + c}}$

• Objects with smaller moments of inertia divert a smaller fraction of potential energy into rotational kinetic energy, leaving more energy available for translational motion.

• Arrival order at the bottom of the ramp (fastest to slowest):

  1. Solid sphere ($v \approx 1.195\sqrt{gh}$)

  2. Solid disk ($v \approx 1.155\sqrt{gh}$)

  3. Spherical shell ($v \approx 1.095\sqrt{gh}$)

  4. Hoop ($v = 1.000\sqrt{gh}$)

Lecture Questions and Concept Checks

• Question 1: Cable Tension in Static Equilibrium:

  • Prompt: Use the equilibrium condition to solve for the tension $T$ in the rope supporting the $2.00\,\text{kg}$ mass on the beam (to 3 significant figures).

  • Derivation:     $\sum \tau = (0.500\,\text{m})T\sin(80^\circ) - (2.50\,\text{m})(2.00\,\text{kg})(9.8\,\text{m/s}^2) = 0$     $T\sin(80^\circ) = 98.0\,\text{N} \implies T = \frac{98.0\,\text{N}}{\sin(80^\circ)} = 99.5\,\text{N}$

  • Answer: $99.5\,\text{N}$

• Question 2: Pivot Normal Reaction Force Components:

  • Prompt: Find the components of the reaction force $\vec{N}$ exerted by the wall pivot, starting from $\vec{F}_{\text{net}} = m\vec{a} = 0$.

  • Derivation:

  • $N_x = -T_x = -(-T\cos(80^\circ)) = 99.5\,\text{N} \cdot \cos(80^\circ) = 17.3\,\text{N}$

  • $N_y = -T_y + mg = -98.0\,\text{N} + (2.00\,\text{kg})(9.8\,\text{m/s}^2) = -78.4\,\text{N}$

  • Answer: $N_x = 17.3\,\text{N}$, $N_y = -78.4\,\text{N}$

• Question 3: Angular Velocity of a Rotating Disc:

  • Prompt: Suppose it takes $2\,\text{s}$ for this disc to complete a revolution. What is the disc\'s angular velocity $\omega$?

  • Options:

  • A: $1\,\text{rad/s}$

  • B: $2\pi\,\text{rad/s}$

  • C: $\pi\,\text{rad/s}$

  • D: $2\,\text{rad/s}$

  • Derivation:     $\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi\,\text{rad}}{2\,\text{s}} = \pi\,\text{rad/s}$

  • Answer: C ($\pi\,\text{rad/s}$)

• Question 4: Angular Acceleration in Uniform Circular Motion:

  • Prompt: When an object is in uniform circular motion, what is the angular acceleration $\alpha$?

  • Options:

  • A: $0$

  • B: $\frac{v^2}{r}$

  • C: $\omega^2 r$

  • D: $\omega^2$

  • Reasoning:

  • In uniform circular motion, the angular speed $\omega = \frac{v}{r}$ is constant over time.

  • By definition, $\alpha = \frac{d\omega}{dt} = 0$. The inward acceleration is purely centripetal ($a_c = \frac{v^2}{r} = \omega^2 r$), not tangential or angular.

  • Answer: A ($0$)

• Question 5: Torque Dependence on Radial Distance:

  • Prompt: System A has a mass $m$ orbiting at radius $2r$. System B has an identical mass $m$ orbiting at radius $r$. Both undergo identical angular acceleration $\alpha$. Which one requires more torque? How much more?

  • Derivation:     $\tau_A = m(2r)^2\alpha = 4mr^2\alpha$     $\tau_B = mr^2\alpha$     $\frac{\tau_A}{\tau_B} = 4$

  • Answer: System A requires $4\times$ more torque than System B.

• Question 6: Total Torque on a Three-Mass Assembly (End Pivot):

  • Prompt: What is the total torque on the "shish kebab / dango" object consisting of three point masses $m$ located at distances $0$, $r$, and $2r$ from the axis of rotation for an angular acceleration $\alpha$?

  • Options:

  • A: $\tau = 2mr^2\alpha$

  • B: $\tau = 3mr^2\alpha$

  • C: $\tau = 4mr^2\alpha$

  • D: $\tau = 5mr^2\alpha$

  • Derivation:     $\tau = m(0)^2\alpha + m(r)^2\alpha + m(2r)^2\alpha = 0 + mr^2\alpha + 4mr^2\alpha = 5mr^2\alpha$

  • Answer: D ($\tau = 5mr^2\alpha$)

• Question 7: Torque Comparison Between End and Center Pivots:

  • Prompt: Which configuration requires more torque: pivoting the three-mass system at one end or pivoting it at the center? How much more?

  • Derivation:

  • End pivot: $\tau_{\text{end}} = 5mr^2\alpha$

  • Center pivot: $\tau_{\text{center}} = m(-r)^2\alpha + m(0)^2\alpha + m(r)^2\alpha = 2mr^2\alpha$

  • Ratio: $\frac{\tau_{\text{end}}}{\tau_{\text{center}}} = \frac{5}{2} = 2.5$

  • Answer: The end-pivot configuration requires more torque ($2.5\times$ as much torque as the center-pivot configuration).

• Question 8: Maximum Rotational Inertia Among Standard Geometries:

  • Prompt: Suppose a hoop, disk, solid sphere, and thin spherical shell each have the same mass $M$ and radius $R$. Which object resists changes to its rotation the most?

  • Options:

  • A: Hoop

  • B: Disk

  • C: Solid sphere

  • D: Spherical shell

  • Reasoning:

  • The rotational inertia coefficients are: Hoop ($1.0$), Spherical shell ($0.67$), Disk ($0.50$), Solid sphere ($0.40$).

  • The hoop has all of its mass situated at the maximum possible distance $R$ from the axis, giving it the largest moment of inertia ($I = MR^2$).

  • Answer: A (Hoop)

• Question 9: Ranking Rotational Resistance:

  • Prompt: Four objects (hoop, disk, solid sphere, shell) of identical mass $M$ and radius $R$ are placed on a ramp with friction at height $h$. Rank the objects from least to most resistant to changes in rotational motion.

  • Ranking:

  1. Least resistant: Solid sphere ($I = 0.40\,MR^2$)

  2. Second least resistant: Disk ($I = 0.50\,MR^2$)

  3. Second most resistant: Thin spherical shell ($I = 0.67\,MR^2$)

  4. Most resistant: Hoop ($I = 1.00\,MR^2$)

• Question 10: Incline Race Winner:

  • Prompt: The four objects all begin rolling down the ramp at the same time from height $h$. Which one reaches the bottom first?

  • Options:

  • A: Hoop

  • B: Disk

  • C: Solid sphere

  • D: Shell

  • Reasoning:

  • The solid sphere has the smallest moment of inertia, converting the least amount of potential energy into rotational kinetic energy.

  • As a result, it achieves the highest translational velocity ($v = \sqrt{\frac{10}{7}gh} \approx 1.195\sqrt{gh}$) and reaches the bottom first.

  • Answer: C (Solid sphere)

• Question 11: Total Kinetic Energy Comparison:

  • Prompt: Which object has the most total kinetic energy ($KE$) at the bottom of the ramp?

  • Options:

  • A: Hoop

  • B: Disk

  • C: Solid sphere

  • D: All have the same $KE$

  • Reasoning:

  • Conservation of mechanical energy states that $KE_f = PE_i = mgh$.

  • Because all four objects have identical mass $m$ and start from the same height $h$, each object converts the exact same amount of gravitational potential energy into total mechanical kinetic energy ($KE_{\text{trans}} + KE_{\text{rot}}$).

  • Answer: D (All have the same $KE$)