| 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. |
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A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom?
\( \text{Solid sphere: } I = \frac{2}{5}mR^2, \quad \text{Solid disk: } I = \frac{1}{2}mR^2, \quad \text{Hoop: } I = mR^2 \)
To increase the moment of inertia of a body about an axis, you must
Two uniform disks have the same radius but different masses: disk \( 1 \) has a mass \( M \), disk \( 2 \) has a mass \( 2M \). What is the ratio of the moment of inertia of the first disk to the second disk?
A solid metal bar is at rest on a horizontal frictionless surface. It is free to rotate about a vertical axis at the left end. The figures below show forces of different magnitudes that are exerted on the bar at different locations. In which case does the bar’s angular speed about the axis increase at the fastest rate?
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously from the top of an inclined plane and do not slip, which one will reach the bottom first?
To increase the moment of inertia of a body about an axis, you must
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?
Which of the following must be zero if an object is spinning at a constant rate? There may be more than one right answer.
An old record player could bring a disk up to its \(45\) RPM speed in less than a second. If the same size disk can also be brought up to a speed of \(75\) RPM in about the same amount of time on another player. Compare the torques exerted by each record player.
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first? Hint the rotational inertia of a sphere about its center is \(I = \frac{2}{5}mR^{2}\) and the rotational inertia of a cylinder about its center is \(I = \frac{1}{2}mR^{2}\).
\(1.87\times 10^{1}\,\text{kg m}^{2}\)
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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] |
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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