| Derivation/Formula | Reasoning |
|---|---|
| \[I_0 = \tfrac{1}{2} M R^{2}\] | Initial moment of inertia \(I_0\) for a solid cylinder with mass \(M\) and radius \(R\). |
| \[I_a = \tfrac{1}{2} M (2R)^{2} = 2 M R^{2}\] | Option (a): Doubling the radius while \(M\) is constant gives \(I_a = 4 I_0\); the increase is \(3 I_0\). |
| \[I_b = \tfrac{1}{2} (4M) R^{2} = 2 M R^{2}\] | Option (b): Quadrupling the mass while \(R\) is constant also yields \(I_b = 4 I_0\); the increase is \(3 I_0\). |
| \[I_c = \tfrac{1}{2} (2M) R^{2} = M R^{2}\] | Option (c): Doubling \(M\) only doubles the inertia, \(I_c = 2 I_0\); the increase is \(I_0\), therefore not the same. |
| \[I_d = \tfrac{1}{2} M (4R)^{2} = 8 M R^{2}\] | Option (d): Quadrupling \(R\) makes \(I_d = 16 I_0\); the increase is \(15 I_0\), therefore not the same. |
| \[\boxed{(a)\,\&\,(b)}\] | Both (a) and (b) increase the moment of inertia by exactly the same amount because they each produce a four-fold final value. Options (c) and (d) alter the inertia by different factors, so they are incorrect. |
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The moment of inertia of a uniform solid sphere (mass \( M \), radius \( R \)) about a diameter is \( \frac{2}{5}MR^2 \). The sphere is placed on an inclined plane (angle \( \theta \)) and released from rest.
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(2M\) that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?
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?
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
A Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
A \( 0.72 \) \( \text{m} \)-diameter solid sphere can be rotated about an axis through its center by a torque of \( 10.8 \) \( \text{Nm} \) which accelerates it uniformly from rest through a total of \( 160 \) revolutions in \( 15.0 \) \( \text{s} \). What is the mass of the sphere?
To increase the moment of inertia of a body about an axis, you must
A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?
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?
Two masses, \( m_y = 32 \) \( \text{kg} \) and \( m_z = 38 \) \( \text{kg} \), are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius \( R = 0.311 \) \( \text{m} \) and mass \( 3.1 \) \( \text{kg} \). Initially, \( m_y \) is on the ground and \( m_z \) rests \( 2.5 \) \( \text{m} \) above the ground.
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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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