Chapter 7–10 Notes: Energy, Work, Conservation, Momentum, and Rotation
7.1 Systems and Environments
A system is a selected portion of the universe that we study; the environment is everything outside the system. A system boundary is an imaginary surface that divides the system from its environment. The boundary can be a real surface or an arbitrary surface chosen for analysis.
Valid systems can be a single object, a collection of objects, a region of space, or even deformable bodies (e.g., a rubber ball that deforms upon impact). The Categorize step in problem solving involves choosing the system, and the boundary defines what the environment can influence.
Examples: a chalkboard eraser moving on a tray; the combination of a ball, a block, and a cord in a pulley system. External influences include gravity, normal forces, friction, and forces transmitted through cords or pulleys. Internal forces (e.g., forces between components that are inside the system) are not counted as environmental influences.
Work is one mechanism by which the environment can influence a system and transfer energy across the system boundary.
7.2 Work Done by a Constant Force
Work is defined for a constant force as the projection of the displacement onto the force:
W = F\cdot r = Fr\cos u,
where F is the force magnitude, r is the displacement magnitude, and u is the angle between the force and displacement vectors.Work is a scalar, even though it is defined by two vectors. It can be positive or negative depending on whether the force component along the displacement is in the same or opposite direction.
Perpendicularity (u = 90°) gives W = 0; no displacement along the force direction means no work.
The sign of work depends on the projection of F onto r; e.g., lifting an object: the applied force and the displacement are in the same direction, so W > 0; gravity does negative work in lifting because its direction opposes the displacement.
The unit of work is the joule: 1\ \text{J} = 1\ \text{N}\cdot\text{m} = \text{kg} \cdot \text{m}^2/\text{s}^2.
Work is an energy transfer. If W > 0, energy is transferred to the system; if W < 0, energy is transferred from the system.
7.3 The Scalar Product of Two Vectors
The work expression can be written using the scalar (dot) product:
W = \mathbf{F}\cdot \mathbf{r} = Fr\cos u.The scalar product is commutative:
{\bf A}\cdot{\bf B} = {\bf B}\cdot{\bf A}.For unit vectors, the dot product reduces to the sum of products of components:
\mathbf{A}\cdot\mathbf{B} = AxBx + AyBy + AzBz.The dot product relates to projections: the dot product equals the magnitude of one vector times the projection of the other onto that vector.
Work remains a scalar even when F and r are vectors.
7.4 Work Done by a Varying Force
When the force varies with position, the work is the sum (integral) of the small displacements:
W = \int{xi}^{xf} Fx\,dx,
where Fx is the x-component of the force. This corresponds to the area under the Fx vs x curve.For a net force acting along the x-axis with displacement from xi to xf, the work is
W = \int{\xi}^{\xf} Fx\,dx.If multiple forces act, the net work is the integral of the net force along the path:
W{net} = \int{\xi}^{\xf} \mathbf{F}_{net}\cdot d\mathbf{r}.For non-particle (deformable) systems, forces might act across different internal displacements; the internal forces may be treated as internal to the system and not included in the environment’s work.
Example interpretations include a force that is constant for the first portion of motion and then varies, with the total work equal to the area under the force curve (the sum of rectangle and triangle areas in common diagrams).
7.5 Work Is a Scalar
Although work is defined via vectors, it is a scalar quantity. This is advantageous because energy methods avoid vector algebra when applying the energy approach.
This scalar nature is consistent with energy being a scalar quantity and all energy transfers (kinetic, potential, internal) being scalars.
7.6 Potential Energy of a System
Potential energy (Us) is stored energy associated with the configuration of a system, typically arising from conservative forces.
Gravitational potential energy near the surface:
U_g = mgy,
where y is measured from a chosen reference level. The change in Ug depends only on vertical displacement, not on the path taken.Elastic (spring) potential energy:
Us = \frac{1}{2} kx^2, where x is the displacement from the spring's equilibrium (x = 0 at un-stretched length). The restoring force is conservative and equals $Fs = -\dfrac{dU_s}{dx}$.Potential energy is associated with a system of interacting objects; Earth is included in the system for gravitational potential energy problems because the gravitational force acts between system members.
7.7 Conservative and Nonconservative Forces
Conservative forces have two key properties:
1) The work done by a conservative force between two points is independent of the path taken.
2) The work done by a conservative force around any closed path is zero.Gravitational and elastic (spring) forces are conservative.
Nonconservative forces (e.g., kinetic friction) do not satisfy these properties; their work depends on the path and cannot be expressed purely as a potential energy difference.
Energy accounting: The total mechanical energy is the sum of kinetic and potential energies, and is conserved for isolated systems with only conservative forces acting. Nonconservative forces can convert mechanical energy into internal energy (e.g., heat).
7.8 Relationship Between Conservative Forces and Potential Energy
For a conservative force, the work done between two configurations is the negative change in the corresponding potential energy:
Wc = Ui - U_f \quad \text{or} \quad dU = -\mathbf{F}\cdot d\mathbf{r}.The force can be obtained from the potential energy as the negative gradient:
\mathbf{F} = -\nabla U.Examples: for a spring, $F=-k x$ and $Us=\tfrac{1}{2}kx^2$; for gravity near the Earth’s surface, $Ug = m g y$ and $F_g = -m g \hat{j}$ along -y.
The change in potential energy associated with a conservative force is related to the work done by that force: $W_c = -\Delta U$.
7.9 Energy Diagrams and Equilibrium of a System
A graph of Us versus a configuration coordinate (e.g., x for a spring) illustrates energy landscapes.
Equilibrium points correspond to where the force vanishes, i.e., where the slope of U is zero. Stability is determined by the curvature: a minimum U corresponds to stable equilibrium; a maximum U to unstable equilibrium; a flat region corresponds to neutral equilibrium.
The Lennard–Jones potential is an example of a molecular-scale Us(x) with a minimum at the equilibrium separation; the minimum corresponds to a stable equilibrium position.
8.1 The Nonisolated System: Conservation of Energy
Energy can cross the system boundary via several transfer mechanisms: work (W), heat (Q), mechanical waves (TMW), matter transfer (TMT), electrical transmission (TET), and electromagnetic radiation (TER).
The total energy of a nonisolated system evolves as:
E{ ext{system}} = K + U + E{ ext{int}} = W + Q + T{ ext{MW}} + T{ ext{MT}} + T_{ ext{ET}} + TER.The general conservation of energy equation is:
\Delta E{ ext{system}} = E{ ext{transfers}}.The special case of the work–kinetic energy theorem is recovered when the only transfer is work across the boundary: \Delta K = W.
8.2 The Isolated System
For an isolated system, no energy crosses the boundary: \Delta E_{ ext{system}} = 0.
If only conservative forces act within the isolated system, the mechanical energy is conserved: \Delta E{ ext{mech}} = \Delta(K + U) = 0. or Kf + Uf = Ki + U_i.
In a gravitational system, this leads to Ug + K conservation: \Delta K = -\Delta U_g.
If nonconservative forces act within an isolated system, the total energy Esystem is conserved, but mechanical energy Met (K+U) may change into internal energy: \Delta E{ ext{system}} = 0,\quad E_{ ext{int}} \text{ can increase}.
8.3 Situations Involving Kinetic Friction
For a block on a surface with kinetic friction, the work–kinetic energy relation is modified because friction does work on the system and converts kinetic energy into internal energy:
\Delta K = W{ ext{other forces}} - fk d,
where d is the displacement along the path and f_k is the kinetic friction force.The total energy change of the system includes the increase in internal energy due to friction: \Delta E{ ext{int}} = fk d.
In an isolated system with friction, the mechanical energy decreases while internal energy increases; in a nonisolated system, the work done by external forces plus friction accounts for the energy changes.
8.4 Changes in Mechanical Energy for Nonconservative Forces
When nonconservative forces act within a system, the change in mechanical energy is given by
\Delta E{ ext{mech}} = - (fk d) + W_{ ext{other forces}}.If friction is present along an incline, some energy is transformed into internal energy, while gravity can still decrease the gravitational potential energy, contributing to total energy changes.
Problem-solving strategy for nonconservative systems: identify the system, decide if it is isolated, account for all energy transfer mechanisms, and apply the generalized energy balance accordingly.
8.5 Power
Power is the rate of energy transfer:
P = \frac{dE}{dt} = \frac{dW}{dt}.If a constant external force moves a body with velocity v, average power over a short interval is P = \mathbf{F}\cdot \mathbf{v}.
The unit of power is the watt: 1\ \text{W} = 1\ \text{J/s}.
Other units include horsepower (hp) with 1\text{ hp} = 746\ \text{W}.
9.1 Linear Momentum and Its Conservation
Linear momentum is defined as\mathbf{p}=m\mathbf{v}. For a system of particles, the total momentum is the sum of the individual momenta: \mathbf{p}{\text{tot}} = \sumi mi \mathbf{v}i = M \mathbf{v}{\text{CM}}, where $\mathbf{v}{\text{CM}}$ is the velocity of the center of mass and $M$ is the total mass.
Momentum is a vector; momentum conservation applies to an isolated system, i.e., when the net external force is zero.
Example: A 60 kg archer on frictionless ice fires a 0.50 kg arrow at 50 m/s. Horizontal momentum before equals zero; after firing, $p{1f} + p{2f} = 0$, giving the recoil velocity of the archer:
v{1f} = -\frac{m2}{m1} v{2f}.
9.2 Impulse and Momentum
The impulse of a net force acting over a time interval is
\mathbf{I} = \int{ti}^{tf} \mathbf{F}{\text{net}} dt.Momentum change equals impulse:
\Delta \mathbf{p} = \mathbf{p}f - \mathbf{p}i = \mathbf{I}.The time-averaged net force over the interval relates to impulse by
\mathbf{I} = \mathbf{F}_{\text{avg}} \Delta t.Impulse is a vector with dimensions of momentum, and it represents the transfer of momentum across a system boundary.
9.3 Collisions in One Dimension
Perfectly inelastic collision (objects stick together):
m1 v{1i} + m2 v{2i} = (m1 + m2) v_f.Final velocity for the combined mass:
vf = \frac{m1 v{1i} + m2 v{2i}}{m1 + m_2}.Elastic collision in 1D: momentum and kinetic energy are conserved. Final velocities satisfy:
v{1f} = \frac{m1 - m2}{m1 + m2} v{1i} + \frac{2 m2}{m1 + m2} v{2i},
v{2f} = \frac{2 m1}{m1 + m2} v{1i} + \frac{m2 - m1}{m1 + m2} v{2i}.Relative velocity reverses: $v{1i}-v{2i} = -(v{1f}-v{2f})$ for elastic collisions.
9.4 Collisions in Two Dimensions
Momentum is conserved in each component (x and y) for an isolated system.
Elastic collisions also conserve kinetic energy: $Ki = Kf$.
When masses and velocities are known, the problem reduces to solving for speeds and directions consistent with momentum conservation in both components and, if elastic, energy conservation.
9.5 The Center of Mass
The center of mass (CM) position is
\mathbf{r}{\text{CM}} = \frac{1}{M} \sumi mi \mathbf{r}i. For continuous mass distributions, this becomes an integral:
\mathbf{r}_{\text{CM}} = \frac{1}{M} \int \mathbf{r}\, dm.The velocity of CM is
\mathbf{v}{\text{CM}} = \frac{1}{M} \sumi mi \mathbf{v}i = \frac{\mathbf{p}_{\text{tot}}}{M}.The CM moves as if all mass were concentrated at CM under the net external force:
\mathbf{F}{\text{ext}} = M \mathbf{a}{\text{CM}}.Impulse on the CM relates to changes in CM momentum:
\Delta (M\mathbf{v}{\text{CM}}) = \int \mathbf{F}{\text{ext}} dt.If the net external force is zero, total momentum is conserved for the system, and CM velocity is constant.
9.6 Motion of a System of Particles
The total momentum change of the system equals the impulse from external forces:
\mathbf{p}{\text{tot}f} - \mathbf{p}{\text{tot}i} = \int{ti}^{tf} \mathbf{F}{\text{ext}} dt.If external force is zero, total momentum is conserved: \mathbf{p}{\text{tot}f} = \mathbf{p}{\text{tot}i}.
9.7 Deformable Systems
Systems that deform (e.g., a person pushing off a wall) require energy and momentum analyses that go beyond the simple particle model.
In such cases, conservation of energy and conservation of momentum can still be applied, but the energy can transform between kinetic and internal forms, and energy transfer must be tracked across the system and boundary.
9.8 Rocket Propulsion
Momentum conservation for a rocket plus expelled fuel in free space yields the rocket equation:
\frac{d}{dt}(M v) = -ve \frac{dM}{dt}, or, integrated (with exhaust speed $ve$ relative to rocket):
vf = vi + ve \ln\left(\frac{Mi}{M_f}\right).Thrust arises from exhaust mass flow:
FT = ve \frac{dM}{dt}.Examples include rockets firing in space and other propulsion problems; the key idea is momentum exchange with the expelled propellant.
10.1 Angular Position, Velocity, and Acceleration
In rotational motion about a fixed axis, the angular position $\theta$ is measured relative to a reference line.
The angular displacement is
\Delta \theta = \thetaf - \thetai.The average angular speed is
\omega_{\text{avg}} = \frac{\Delta \theta}{\Delta t}The instantaneous angular speed is
\omega = \frac{d\theta}{dt}.The angular acceleration is
\alpha = \frac{d\omega}{dt}.Right-hand rule: the direction of $\boldsymbol{\omega}$ is given by curling the fingers in the direction of rotation; the thumb points along the angular velocity axis.
10.2 Rotational Kinematics: The Rigid Object Under Constant Angular Acceleration
For a rigid object with constant angular acceleration, the kinematic relations are analogous to linear motion:
\omegaf = \omegai + \alpha t,
\thetaf = \thetai + \omegai t + \frac{1}{2}\alpha t^2, \omegaf^2 = \omegai^2 + 2\alpha(\thetaf - \theta_i).Similar to translation, there are also expressions relating angular displacement, velocity, and time for constant angular acceleration.
10.3 Angular and Translational Quantities
Tangential (linear) speed of a point at distance $r$ from the axis with angular speed $\omega$:
v = \omega r.Tangential acceleration: $a_t = \alpha r$.
Centripetal (radial) acceleration: $a_c = \dfrac{v^2}{r} = \omega^2 r$.
The total acceleration of a point on the rotating object has tangential and radial components: $a^2 = at^2 + ac^2$.
For rolling without slipping, the center of mass velocity is related to angular speed by
v{\text{CM}} = \omega R, and the CM acceleration by a{\text{CM}} = \alpha R.
10.4 Rotational Kinetic Energy
The kinetic energy of a rotating rigid object is
K_R = \frac{1}{2} I \omega^2,
where $I$ is the moment of inertia about the rotation axis.If the object rotates about a fixed axis with many mass elements, the total rotational KE is
KR = \frac{1}{2} \omega^2 \sumi mi ri^2 = \frac{1}{2} I \omega^2.The moment of inertia $I$ has dimensions of $\text{ML}^2$ and depends on how mass is distributed relative to the rotation axis; the axis of rotation matters (no single $I$ for a given object).
10.5 Calculation of Moments of Inertia
The moment of inertia for a continuous distribution is
I = \int r^2\,dm. For a uniform density, $dm = \rho dV$ (or use $dm = \rho A\,dx$ for simpler geometries).Common geometries and axes have standard $I$ values (Table 10.2). The parallel-axis theorem relates the moment of inertia about any axis to that about a parallel axis through the CM:
I = I_{CM} + M d^2.Examples include a uniform rod, solid cylinder, solid sphere, hollow cylinder, etc., with $I$ values depending on geometry and axis.
10.6 Torque
Torque about a rotation axis is defined as
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
with magnitude
\tau = r F \sin\phi.The moment arm $d$ is the perpendicular distance from the rotation axis to the line of action of the force: $d = r\sin\phi$.
Torque depends on your choice of axis; different axes yield different torques for the same force.
10.7 The Rigid Object Under a Net Torque
The rotational analog of Newton’s second law for a rigid body rotating about a fixed axis is
\boldsymbol{\tau}_{\text{net}} = I \boldsymbol{\alpha}.For a point or small mass element $dm$ at distance $r$, the torque is $d\tau = r\, dFt = r at dm$; integrating gives
\tau_{\text{net}} = I \alpha.The net torque acts on the entire rigid body; the angular acceleration is proportional to the net torque with $I$ as the constant of proportionality.
10.8 Energy Considerations in Rotational Motion
Work done by a torque during a small rotation $d\theta$ is
dW = \tau\, d\theta.The rate of work (power) is
P = \tau \omega.The rotational work–kinetic energy theorem states that the net work done by external forces equals the change in rotational kinetic energy:
W{net} = \Delta KR = \frac{1}{2} I \omegaf^2 - \frac{1}{2} I \omegai^2.For a rolling rigid body, the total kinetic energy is the sum of the translational and rotational kinetic energies:
K = \frac{1}{2} M v{\text{CM}}^2 + \frac{1}{2} I{CM} \omega^2.The rolling relation between translational and rotational motion is $v_{\text{CM}} = \omega R$ for pure rolling, leading to $K = \tfrac{1}{2} I \omega^2$ when expressed about the instantaneous axis through the point of contact.
10.9 Rolling Motion of a Rigid Object
For pure rolling, $v{\text{CM}} = \omega R$ and $a{\text{CM}} = \alpha R$.
The total kinetic energy for a rolling object is the sum of translational and rotational parts:
K = \frac{1}{2} M v{\text{CM}}^2 + \frac{1}{2} I{CM} \omega^2 = \frac{1}{2} I \omega^2 + \frac{1}{2} M (\omega R)^2.The parallel-axis theorem allows expressing $I$ about the CM or about the contact axis to connect rotational and translational energies.
Energy methods can be used to analyze rolling on inclines, and rolling friction converts some mechanical energy to internal energy, though rolling without slipping is often energy-conserving if friction is static and does not dissipate energy.
Example problems illustrate that the acceleration and final speeds for rolling solids depend on the moment of inertia and are generally different from purely sliding objects: for a given incline, a rolling solid typically accelerates more slowly than a sliding block because part of the energy goes into rotation.
Quick Reference: Key Equations (selected)
Work by a constant force: W = \mathbf{F}\cdot \Delta \mathbf{r} = F\Delta r\cos u.
Dot product: \mathbf{A}\cdot\mathbf{B} = AB\cos\theta.
Work by a varying force: W = \int{xi}^{xf} Fx\,dx.
Kinetic energy: K = \tfrac{1}{2}mv^2.
Work–Kinetic Energy Theorem (one dimension): W_{net} = \Delta K.
Potential energy (gravitational): U_g = mgy.
Spring potential energy: U_s = \tfrac{1}{2}kx^2.
Conservative force relation: F = -\frac{\mathrm{d}U}{\mathrm{d}x}.
Total mechanical energy (isolated): E{mech} = K + U, \quad E{tot} = E{mech} + E{int}.
Nonconservative energy change: \Delta E{mech} = -fk d + W{other}, with internal energy increasing by $fk d$ for frictional dissipation.
Power: P = \frac{dE}{dt} = \mathbf{F}\cdot \mathbf{v} = \frac{dW}{dt}.
Linear momentum: \mathbf{p} = m\mathbf{v}, and total momentum conservation: \mathbf{p}_{\text{tot}} = \text{constant}.
Impulse–momentum theorem: \Delta \mathbf{p} = \mathbf{I} = \int \mathbf{F}_{\text{net}} dt.
Elastic collision (1D): velocity relations as given above; relative velocity reverses: $v{1i}-v{2i} = -(v{1f}-v{2f})$.
Center of mass relation: \mathbf{p}{\text{tot}} = M \mathbf{v}{\text{CM}}, \quad \mathbf{F}{\text{ext}} = M \mathbf{a}{\text{CM}}.
Rocket propulsion: \frac{dv}{dt} = -ve \frac{1}{M}\frac{dM}{dt} \quad \Rightarrow\quad vf = vi + ve \ln\left(\frac{Mi}{Mf}\right).
Torque: \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}, \quad |\tau| = rF\sin\phi.
Rotational equation: \boldsymbol{\tau}_{\text{net}} = I\boldsymbol{\alpha}.
Rotational kinetic energy: K_R = \tfrac{1}{2}I\omega^2.
Rolling motion (pure rolling): v{\text{CM}} = \omega R, \quad a{\text{CM}} = \alpha R.
Moment of inertia (definition): I = \int r^2\,dm.
Parallel-axis theorem: I = I_{CM} + Md^2.
Energy in rotation: W = \int \boldsymbol{\tau}\, d\theta = \int \tau \, d\theta = \Delta K_R.
Power in rotation: P = \tau\omega.
Energy balance for rolling on inclines: the total energy is conserved in the absence of nonconservative forces; friction converts some energy to internal forms if rolling friction is present.