| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ K_{\text{trans}} = \tfrac{1}{2} M_{\text{tot}} v^{2} \] | The wagons have the same total mass and reach the same speed, so each gains identical translational kinetic energy. |
| 2 | \[ K_{\text{rot}} = \tfrac{1}{2} I \omega^{2}, \qquad \omega = \frac{v}{R} \] | Rotational kinetic energy of a wheel rolling without slipping; the angular speed is \(\omega=v/R\). |
| 3 | Solid disk: \[ I_{\text{disk}} = \tfrac{1}{2} M_{w} R^{2}\] \[ K_{\text{rot,disk}} = \tfrac{1}{2}\left(\tfrac{1}{2}M_{w}R^{2}\right)\left(\tfrac{v}{R}\right)^{2}=\tfrac{1}{4}M_{w}v^{2} \] | Substitute the disk’s moment of inertia into the rotational energy formula. |
| 4 | Hollow hoop: \[ I_{\text{hoop}} = M_{w} R^{2}\] \[ K_{\text{rot,hoop}} = \tfrac{1}{2}\left(M_{w}R^{2}\right)\left(\tfrac{v}{R}\right)^{2}=\tfrac{1}{2}M_{w}v^{2} \] | The hoop’s larger moment of inertia doubles the rotational energy (for the same mass) compared with a disk. |
| 5 | \[ K_{\text{total,wheel}} = K_{\text{trans,wheel}} + K_{\text{rot}} = \tfrac{1}{2}M_{w}v^{2}+K_{\text{rot}} \] | Each wheel possesses both translational and rotational kinetic energy. |
| 6 | Disk wheel: \[ K_{\text{total,disk}} = \tfrac{1}{2}M_{w}v^{2}+\tfrac{1}{4}M_{w}v^{2}=\tfrac{3}{4}M_{w}v^{2} \] | The extra energy beyond pure translation is \(\Delta K_{\text{disk}} = \tfrac{1}{4}M_{w}v^{2}\). |
| 7 | Hoop wheel: \[ K_{\text{total,hoop}} = \tfrac{1}{2}M_{w}v^{2}+\tfrac{1}{2}M_{w}v^{2}=M_{w}v^{2} \] | The extra energy for a hoop is \(\Delta K_{\text{hoop}} = \tfrac{1}{2}M_{w}v^{2}\). |
| 8 |
Wagon A (disk, \(M_{w}=0.5\,\text{kg}\)): \[ \Delta K_{A}=4\left(\tfrac{1}{4}\times0.5\,v^{2}\right)=0.5\,v^{2} \] Wagon B (disk, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{B}=4\left(\tfrac{1}{4}\times0.2\,v^{2}\right)=0.2\,v^{2} \] Wagon C (hoop, \(M_{w}=0.1\,\text{kg}\)): \[ \Delta K_{C}=4\left(\tfrac{1}{2}\times0.1\,v^{2}\right)=0.2\,v^{2} \] |
Multiply the extra energy per wheel by four wheels for each wagon, using the correct wheel masses. |
| 9 | With \(v=10\,\text{m/s}\) (so \(v^{2}=100\)): \[ \Delta K_{A}=0.5\times100=50\,\text{J} \] \[ \Delta K_{B}=0.2\times100=20\,\text{J} \] \[ \Delta K_{C}=0.2\times100=20\,\text{J} \] |
Compute the numerical extra rotational energy required for each wagon’s wheels. |
| 10 | \[\boxed{\text{Wagon A}}\] | All wagons share the same translational energy, but Wagon A has the largest additional wheel energy (50 J). Hence it needs the most total energy input. |
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A massless rigid rod of length [katex]3d[/katex] is pivoted at a fixed point [katex]W[/katex], and two forces each of magnitude [katex]F[/katex] are applied vertically upward as shown above. A third vertical force of magnitude [katex]F[/katex] may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude [katex]F[/katex] is applied at point?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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