Unit 22-30 Mechanics: Energy, Momentum, and Rotational Dynamics

Chapter 9: Work, Kinetic Energy, and Power

Power Definition: Power is the rate at which energy is transformed or transferred. It is mathematically defined as the derivative of system energy with respect to time: P = \frac{dE_{sys}}{dt}
Mechanical Power and Work-Rate: In terms of work ( $W$ ), power is the rate at which work is done: dW = \mathbf{F} \cdot d\mathbf{r} P = \frac{dW}{dt} = \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = \mathbf{F} \cdot \mathbf{v} = Fv \cos(\phi)
Units and Conversions: - Unit: $J/s = \text{Watts (W)}$ - A common unit is the horsepower ( $hp$ ).
Examples of Power Output: - Lightbulb (100 W): Transforms 100 J of electric energy into light and thermal energy every second. - Athlete (350 J/s or ~0.5 hp): Transforms chemical energy of glucose and fat into mechanical energy at a rate of $350\,J/s$ . - Gas Furnace (20 kW): Transforms chemical energy of gas into thermal energy at a rate of $20,000\,J/s$ .
EXAMPLE 9.11: Power output of a motor: - Scenario: A motor and cable drag a $300\,kg$ machine across a factory floor at a speed of $0.50\,m/s$ . The coefficient of friction ( $\mu_k$ ) is $0.60$ . - Calculation: - Normal force $n = mg = (300)(9.8) = 2940\,N$ . - Friction force $f_k = \mu_k n = (0.60)(2940) = 1764\,N$ . - Tension $T$ must equal $f_k$ for constant speed ( $a = 0$ ): T = f_k = 1764\,N - Power supplied by the motor: P = \mathbf{F} \cdot \mathbf{v} = Tv = (1764)(0.50) = 882\,W
Example: Lifting a motor: - Scenario: Lift a $35\,kg$ motor $3.0\,m$ in $8.0\,s$ using a rope and pulley. - Calculation: - Weight of motor $mg = (35)(9.81) = 343.35\,N$ . - Work done $W = F \Delta y = mg \Delta y = (35)(9.81)(3.0) = 1029.6\,J$ . - Power $P = \frac{W}{\Delta t} = \frac{1030}{8.0} = 130\,W$ . - Alternative Method: Average speed $v = \frac{\Delta y}{\Delta t} = \frac{3.0}{8.0} = 0.375\,m/s$ . Then $P = Fv = mgv = (35)(9.81)(0.375) = 130\,W$ .

Chapter 10: Interactions and Potential Energy

Potential Energy Fundamentals: What is defined as part of the system impacts what is considered internal work versus external work.
Gravitational Potential Energy ( $U_g$ ): - Defined for an object-Earth system where the book and Earth are both within the system boundary. - Work done by an external force lifting a book increasing the system energy: \Delta E_{syst} = \Delta U_g = W_{ext} dU_g = F_r dy \cos(0) = mg dy - Change in potential energy: U_{gf} - U_{gi} = mgy_f - mgy_i - Absolute potential energy (assuming $U_g = 0$ at $y = 0$ ): U_g = mgy
Elastic Potential Energy ( $U_{sp}$ ): - As a spring is compressed or stretched, the force $F_s$ increases ( $F_s = k\Delta x$ ). - Work done by an external force to stretch/compress a spring: W = \int F_x dx = \int kx dx = \frac{1}{2}kx^2 - Elastic Potential Energy formula: U_{sp} = \frac{1}{2}k(\Delta s)^2
Law of Conservation of Energy: - The total energy $E_{sys} = K + U + E_{th}$ of an isolated system is constant. - Internal transformations (kinetic to potential to thermal) occur, but the sum remains unchanged. - Mechanical Energy Conservation: $E_{mech} = K + U$ is conserved if the system is isolated and nondissipative (no friction).
Energy Conservation Problems Modeling: - System Definition: Ensure no external forces do work. If friction exists, include both surfaces in the system. - Visualization: Use before-and-after pictorial representations and energy bar charts. - Isolated and Nondissipative Systems: K_i + U_i = K_f + U_f - Systems with Friction: K_i + U_i = K_f + U_f + \Delta E_{th} - Where thermal energy increase is $\Delta E_{th} = f_k \Delta s$ .
Table 10.2: Choosing Isolated Systems: - Free fall: Ball + Earth. Forces between them are internal. - Frictionless ramp: Object + Earth. Normal force is perpendicular to motion, so it does no work. - Compressing a spring: Object + Spring. Internal forces do no external work. - Sliding with friction: Block + surface. Kinetic friction forces are internal.

Chapter 10 Examples and Practical Applications

EXAMPLE 10.3: The speed of a sled: - Scenario: Christine runs at $2.0\,m/s$ and hops on a sled at the top of a $5.0\,m$ slippery slope. - Equation: $K_i + U_{gi} = K_f + U_{gf}$ - Calculation: \frac{1}{2}mv_i^2 + mgy_i = \frac{1}{2}mv_f^2 + mgy_f v_f = \sqrt{v_i^2 + 2gy_i} = \sqrt{2.0^2 + 2(9.8)(5.0)} = 10\,m/s
Example: Skateboarding up a ramp: - Scenario: Isabella ( $55\,kg$ ) hits a $6.0\,m$ long, $15^{\circ}$ ramp with coefficient of rolling friction $\mu_r = 0.025$ . Target top speed is $2.5\,m/s$ . - Analysis: - Height change $\Delta y = L \sin(15^{\circ}) = 6.0 \sin(15^{\circ}) = 1.55\,m$ . - Friction force $f_r = \mu_r mg \cos(15^{\circ})$ . - Thermal energy $\Delta E_{th} = f_r L = \mu_r mg L \cos(15^{\circ})$ . - Conservation: $K_i + U_{gi} = K_f + U_{gf} + \Delta E_{th}$ . \frac{1}{2}mv_i^2 + 0 = \frac{1}{2}mv_f^2 + mgy_f + \mu_r mg L \cos(15^{\circ}) - Initial speed found to be $v_i = 6.3\,m/s$ . - Thermal Energy Analysis: - $\Delta E_{th} = \mu_r mg L \cos(15^{\circ}) = (0.025)(55)(9.81)(6.0) \cos(15^{\circ}) = 78\,J$ . - Initial Mechanical Energy ( $K_i$ ) = $1090\,J$ . - Percentage lost to friction: $\frac{78}{1090} \approx 7.2\%$ .
EXAMPLE 10.5: Air-track glider compresses a spring: - Scenario: $500\,g$ glider hits a spring, compressing it $2.7\,cm$ . When suspended vertically, the spring stretches $3.5\,cm$ . - Finding spring constant $k$ : - At rest: $F_s = mg \rightarrow k \Delta y = mg$ . - $k = \frac{mg}{\Delta y} = \frac{(0.50)(9.81)}{0.035} = 140\,N/m$ . - Finding initial speed $v_0$ : - $\frac{1}{2}mv_0^2 = \frac{1}{2}k(\Delta x)^2$ - $v_0 = \sqrt{\frac{k}{m}} \Delta x = \sqrt{\frac{140}{0.50}} (0.027) = 0.45\,m/s$ .
EXAMPLE 10.6: Graphing Spring Constant: - Data: $(M, h)$ pairs: $(50\,g, 2.07\,m)$ , $(100\,g, 1.11\,m)$ , $(150\,g, 0.65\,m)$ , $(200\,g, 0.51\,m)$ . - Energy Equation: $\frac{1}{2}k(\Delta y_i)^2 = mgH$ . Rearranging for height: $H = [\frac{k(\Delta y_i)^2}{2g}] \times (\frac{1}{m})$ . - Slope Analysis: Graph of $h$ vs $1/m$ yields a slope of $0.105$ . - $\text{slope} = \frac{k(\Delta y_i)^2}{2g}$ - $k = \frac{2g \cdot \text{slope}}{(\Delta y_i)^2} = \frac{2(9.8)(0.105)}{(0.04)^2} = 1290\,N/m$ .
Example: Speed of a pendulum: - Scenario: $L = 78\,cm$ , $m = 150\,g$ , angle $= 60^{\circ}$ . - Height calculation: $\Delta y = L - L \cos(60^{\circ}) = 0.78(1 - 0.5) = 0.39\,m$ . - Speed: $v_f = \sqrt{2g\Delta y} = \sqrt{2(9.8)(0.39)} = 2.8\,m/s$ .

Conservative and Non-conservative Forces

Conservative Forces: Forces where work done is path-independent (e.g., gravity, spring force). Objects returning to their starting point lose no kinetic energy.
Non-conservative Forces: Work done is path-dependent (e.g., friction, drag, tension). Result in conversion of mechanical energy to thermal energy ( $E_{th}$ , which cannot be converted back).
Work-Energy Principle Revisited: \Delta K = W_{tot} = W_c + W_{nc} = W_c + W_{diss} + W_{ext} \Delta E_{mech} + \Delta E_{th} = \Delta E_{sys} = W_{ext}

Chapter 11: Momentum

Definitions: - Momentum ( $\mathbf{p}$ ): $\mathbf{p} = m\mathbf{v}$ . - Components: $p_x = mv_x$ ; $p_y = mv_y$ . - Relation to Force: $\mathbf{F} = m\mathbf{a} = m\frac{d\mathbf{v}}{dt} = \frac{d(m\mathbf{v})}{dt} = \frac{d\mathbf{p}}{dt}$ . - Impulse (Change in Momentum): $\Delta \mathbf{p} = \mathbf{p}<em>f - \mathbf{p}_i = \int</em>{t_1}^{t_2} \mathbf{F}(t) dt$ .
Conservation of Momentum: The total momentum $\mathbf{P}$ of an isolated system ( $\mathbf{F}_{net} = 0$ ) is constant. \mathbf{P}_i = \mathbf{P}_f
Collisions: - Inelastic Collisions: Objects stick together and move with a common final velocity ( $v_f$ ). Mechanical energy is not conserved (some converted to heat). m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f - Perfectly Elastic Collisions: Mechanical energy is conserved; objects bounce off without loss of energy (e.g., billiard balls). - If object 2 is at rest ( $v_{2i} = 0$ ): v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i} - Three Cases ( $v_{2i}=0$ ): - $m_1 = m_2$ : Ball 1 stops, Ball 2 moves forward with $v_{1i}$ . - $m_1 \gg m_2$ : Ball 1 continues with original speed, Ball 2 knocked forward at $2v_{1i}$ . - $m_1 \ll m_2$ : Ball 1 bounces off with reverse speed, Ball 2 hardly moves.
Explosions: Particles of a system move apart after internal interaction. If isolated, total momentum is conserved (recoil).

Momentum Examples and Problem Solving

Ex: A glider collision (Graphical): - Scenario: $m_1=250\,g$ , $m_2=500\,g$ (at rest). Graph shows slopes (velocities). - Data: $v_{1i} = 0.75\,m/s$ , $v_{1f} = -0.21\,m/s$ . - Equation: $m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$ - Calculation: $(250)(0.75) = (250)(-0.21) + (500)v_{2f}$ . Result: $v_{2f} = 0.48\,m/s$ .
Ex: Rolling away: - Scenario: Bob ( $75\,kg$ ) runs at cart ( $25\,kg$ ) $8.0\,m$ away. Bob accelerates at $1.0\,m/s^2$ . - Step 1: Find Bob's speed at jump: $v^2 = 2a\Delta x = 2(1.0)(8.0) \rightarrow v = 4.0\,m/s$ . - Step 2: Conservation of momentum: $m_B v_{Bi} = (m_B + m_C) v_f$ . - Result: $v_f = \frac{75(4.0)}{100} = 3.0\,m/s$ .
EXAMPLE 11.6: Recoil: - Scenario: $10\,g$ bullet, $3.0\,kg$ rifle, $500\,m/s$ muzzle velocity. - Calculation: $m_B v_B + m_R v_R = 0$ . $v_R = -\frac{m_B v_B}{m_R} = -\frac{(0.010)(500)}{3.0} = -1.7\,m/s$ .
EXAMPLE 11.7: Radioactivity: - Scenario: $^{238}U$ nucleus ( $238\,u$ ) disintegrates. Fragment ejected at $1.50 \times 10^7\,m/s$ . Daughter nucleus recoils at $-2.56 \times 10^5\,m/s$ . - Equation: $m_1 v_{1f} + m_2 v_{2f} = 0$ where $m_1 + m_2 = 238$ . - Result: Fragment mass $m_2 \approx 4\,u$ , daughter mass $m_1 \approx 234\,u$ .
Ex: Ballistic Pendulum: - Scenario: $10\,g$ bullet embeds in $1200\,g$ wood block on $1.50\,m$ string. Swing angle $= 40^{\circ}$ . - Step 1 (Swing): Mechanical energy to find speed after impact ( $V$ ): - $\frac{1}{2}(m+M)V^2 = (m+M)gL(1-\cos(\theta))$ - $V = \sqrt{2(9.8)(1.5)(1 - \cos(40^{\circ}))} = 2.62\,m/s$ . - Step 2 (Collision): Conservation of momentum: - $m v_i = (m+M)V$ - $v_i = \frac{(0.01 + 1.20)(2.62)}{0.01} = 320\,m/s$ .
Ex: Peregrine falcon strike (2D Momentum): - Scenario: $0.80\,kg$ falcon ( $18\,m/s$ at $45^{\circ}$ down) hits $0.36\,kg$ pigeon ( $9.0\,m/s$ horizontal). - X-momentum: $m_F v_{Fi} \cos(45^{\circ}) + m_P v_{Pi} = (m_F + m_P) v_{fx}$ . - Y-momentum: $-m_F v_{Fi} \sin(45^{\circ}) + 0 = (m_F + m_P) v_{fy}$ . - Calculation: $v_{fx} = 11.6\,m/s$ , $v_{fy} = -8.78\,m/s$ . - Final Speed: $v_f = \sqrt{11.6^2 + 8.78^2} = 15\,m/s$ . - Direction: $\phi = \tan^{-1}(\frac{8.78}{11.6}) = 37^{\circ}$ (below horizontal).

Chapter 12: Rotation of a Rigid Body

Rigid-Body Model: Extended object where all points maintain relative positions. Every point has the same angular velocity ( $\omega$ ) and angular acceleration ( $\alpha$ ).
Rotational Kinematics Review: - $\omega = \frac{d\theta}{dt}$ , $\alpha = \frac{d\omega}{dt}$ . - Tangential Velocity: $v_t = r\omega$ . - Tangential Acceleration: $a_t = r\alpha$ . - Centripetal (Radial) Acceleration: $a_r = r\omega^2 = \frac{v^2}{r}$ .
Center of Mass (COM): The axis about which an unconstrained object rotates if no net force is present. x_{cm} = \frac{1}{M} \sum_{i} m_i x_i
Rotational Kinetic Energy: - The sum of kinetic energies of all particles in the rotating object. K_{rot} = \sum \frac{1}{2}m_i v_i^2 = \sum \frac{1}{2}m_i (r_i \omega)^2 = \frac{1}{2} (\sum m_i r_i^2) \omega^2 - Moment of Inertia ( $I$ ): $\sum m_i r_i^2$ . Rotational equivalent of mass. K_{rot} = \frac{1}{2} I \omega^2
Standard Moments of Inertia (Uniform Density): - Thin rod (center): $\frac{1}{12}ML^2$ - Thin rod (end): $\frac{1}{3}ML^2$ - Cylinder or Disk (center): $\frac{1}{2}MR^2$ - Cylindrical hoop (center): $MR^2$ - Solid Sphere (diameter): $\frac{2}{5}MR^2$ - Spherical shell (diameter): $\frac{2}{3}MR^2$
STOP TO THINK 12.2: A solid cylinder ( $1/2 MR^2$ ) and a cylindrical shell ( $MR^2$ ) with same mass and radius rotate with same $\omega$ . The shell has more kinetic energy because its moment of inertia is larger.

Rotational Dynamics and Torque

Torque ((\tau)): Rotational equivalent of force. The ability of a force to cause rotation. \tau = r F \sin(\phi) - Angle $\phi$ is measured counterclockwise from the radial line. - Sign Convention: Positive for Counterclockwise (CCW), Negative for Clockwise (CW). - Units: $N \cdot m$ .
Interpretations of Torque: - Tangential component approach: $\tau = r F_t$ , where $F_t = F \sin(\phi)$ . - Moment arm ( $d$ ) approach: $\tau = d F$ , where $d = r \sin(\phi)$ .
Net Torque ( $\tau_{net}$ ): \tau_{net} = \sum \tau_i
Gravitational Torque: The entire weight of an object can be treated as acting at the Center of Mass (COM). \tau_{grav} = -Mg x_{cm}
Newton's Second Law for Rotational Motion: - A net torque causes angular acceleration: \alpha = \frac{\tau_{net}}{I}
Example: Rotating rockets: - Scenario: Rocket 1 ( $100,000\,kg$ ) and Rocket 2 ( $200,000\,kg$ ) at ends of a $90\,m$ tunnel. Thrusts are $50,000\,N$ in opposite directions. - COM: $x_{cm} = \frac{200,000(90)}{300,000} = 60\,m$ from Rocket 1. - Moment of Inertia: $I = (100k)(60)^2 + (200k)(30)^2 = 5.4 \times 10^8\,kg \cdot m^2$ . - Net Torque: $\tau = (50k)(60) + (50k)(30) = 4.5 \times 10^6\,N \cdot m$ . - Angular acceleration: $\alpha = \frac{4.5 \times 10^6}{5.4 \times 10^8} = 0.00833\,rad/s^2$ . - Velocity after 30 s: $\omega_f = (0.00833)(30) = 0.25\,rad/s$ .
Constraints for Ropes and Pulleys: - Non-slipping condition: The motion of the rim of the pulley must match the attached object. - $v_{obj} = |\omega| R$ - $a_{obj} = |\alpha| R$
Static Equilibrium: - For an extended object to be at rest: 1. $\mathbf{F}<em>{net} = 0$ ( $\sum F_x = 0$ , $\sum F_y = 0$ ) 2. $\tau</em>{net} = 0$ (Torque must be zero about any pivot point).

Rotation and Equilibrium Examples

EXAMPLE 12.12: Lowering a bucket: - Scenario: $2.0\,kg$ bucket on string around a $1.0\,kg$ , $4.0\,cm$ diameter cylinder. - System Equations: - Cylinder: $\tau = TR = I\alpha = (\frac{1}{2}MR^2)\alpha$ . - Constraint: $a = R\alpha$ . - Combined: $T = \frac{1}{2}Ma$ . - Bucket: $mg - T = ma \rightarrow mg - \frac{1}{2}Ma = ma$ . - Acceleration: $a = g / (1 + M/2m) = 9.8 / (1 + 1/4) = 7.84\,m/s^2$ . - Time to travel 1.0 m: $\Delta y = \frac{1}{2}at^2 \rightarrow t = \sqrt{2(1.0)/7.84} = 0.505\,s$ .
EXAMPLE 12.13: Lifting weights (Biceps): - Scenario: Barbell weight $900\,N$ ( $450\,N$ per arm). Tendon is $4.0\,cm$ from elbow, hand is $35\,cm$ from elbow. - Equilibrium: $\tau_{tendon} - \tau_{barbell} = 0$ . - Result: $F_{tendon}(0.04) = (450)(0.35) \rightarrow F_{tendon} = 3900\,N$ .
EXAMPLE 12.15: Will the ladder slip?: - Scenario: $3.0\,m$ ladder, $60^{\circ}$ angle on ground, frictionless wall. - Static Equations: - $\sum F_x = n_2 - f_s = 0$ - $\sum F_y = n_1 - Mg = 0$ - $\sum \tau = Mg \sin(30^{\circ})(1.5) - n_2 \sin(60^{\circ})(3) = 0$ - Solving: $n_2 = \frac{1.5 Mg \sin(30^{\circ})}{3 \sin(60^{\circ})} = \frac{1}{2} Mg \tan(30^{\circ})$ . - Minimum Coefficient: $f_s = n_2 = \mu_s n_1 \rightarrow \mu_s Mg = \frac{1}{2} Mg \tan(30^{\circ})$ . - Result: $\mu_s = \frac{1}{2} \tan(30^{\circ}) \approx 0.29$ .

Rolling Motion

Rolling without slipping condition: $v_{cm} = \omega R$ . After one full rotation, $\Delta x_{cm} = 2\pi R$ .
Total Kinetic Energy for Rolling Object: K = K_{trans} + K_{rot} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 - If $I = cMR^2$ : $K = \frac{1}{2} M v^2 (1 + c)$ .
The Great Downhill Race: - Comparing Sphere ( $c=0.4$ ), Cylinder ( $c=0.5$ ), Hoop ( $c=1.0$ ), and frictionless Particle ( $c=0$ ). - Speed at bottom: $v = \sqrt{\frac{2gh}{1+c}}$ . - Winning Order: Particle (fastest) > Solid Sphere > Solid Cylinder > Circular Hoop (slowest).