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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[I_{\text{cyl}} = \tfrac{1}{2} M_{\text{cyl}} R_{\text{cyl}}^{2}\] | Solid cylinder: use \(\tfrac{1}{2}MR^{2}\). Here \(M_{\text{cyl}} = 100\,\text{kg}\) and \(R_{\text{cyl}} = 0.25\,\text{m}\). |
| 2 | \[I_{\text{cyl}} = 0.5 \times 100 \times 0.25^{2} = 3.125\,\text{kg m}^{2}\] | Numerical substitution to find the cylinder’s inertia. |
| 3 | \[I_{\text{rod (one)}} = \tfrac{1}{3} M_{\text{rod}} L^{2}\] | Each rod is a thin rod about one end (axis at center of wheel). Parallel–axis changes \(\tfrac{1}{12}ML^{2}\) to \(\tfrac{1}{3}ML^{2}\). |
| 4 | \[I_{\text{rod (one)}} = \tfrac{1}{3} \times 5.00 \times 0.75^{2} = 0.9375\,\text{kg m}^{2}\] | Insert \(M_{\text{rod}} = 5.00\,\text{kg}\) and \(L = 0.75\,\text{m}\). |
| 5 | \[I_{\text{4 rods}} = 4 I_{\text{rod (one)}} = 4 \times 0.9375 = 3.75\,\text{kg m}^{2}\] | There are four identical rods. |
| 6 | \[I_{\text{ring}} = M_{\text{ring}} R_{\text{ring}}^{2} = 20.0 \times 0.75^{2} = 11.25\,\text{kg m}^{2}\] | Thin hoop about its center: \(I = MR^{2}\). |
| 7 | \[I_{\text{chunk}} = m_{\text{chunk}} R_{\text{ring}}^{2} = 1.00 \times 0.75^{2} = 0.5625\,\text{kg m}^{2}\] | Treat the metal chunk as a point mass at the ring’s radius. |
| 8 | \[I_{\text{total}} = I_{\text{cyl}} + I_{\text{4 rods}} + I_{\text{ring}} + I_{\text{chunk}}\] | Add the contributions of all parts because they share the same rotational axis. |
| 9 | \[I_{\text{total}} = 3.125 + 3.75 + 11.25 + 0.5625 = 1.86875 \times 10^{1}\,\text{kg m}^{2}\] | Sum of numerical values. |
| 10 | \[\boxed{I_{\text{total}} \approx 1.87 \times 10^{1}\,\text{kg m}^{2}}\] | Moment of inertia of entire assembly, expressed with two significant figures. |
Just ask: "Help me solve this problem."
A \(350\ \text{g}\) ball is attached to the end of a thin, uniform rod of mass \(500\ \text{g}\) and length \(1.2\ \text{m}\). The system is rotated in a horizontal circle about the opposite end of the rod. Calculate the moment of inertia of the system about the axis of rotation. Hint: the moment of inertia of a thin rod about the end of the rod is \(I = \tfrac{1}{3} m L^2\).
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
A spinning ice skater on extremely smooth ice is able to control the rate at which she rotates by pulling in her arms. Which of the following statements are true about the skater during this process?
Three masses are attached to a \( 1.5 \, \text{m} \) long massless bar. Mass 1 is \( 2 \, \text{kg} \) and is attached to the far left side of the bar. Mass 2 is \( 4 \, \text{kg} \) and is attached to the far right side of the bar. Mass 3 is \( 4 \, \text{kg} \) and is attached to the middle of the bar. At what distance from the far left side of the bar can a string be attached to hold the bar up horizontally?
\(1.87\times 10^{1}\,\text{kg m}^{2}\)
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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] |
General Metric Conversion Chart
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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