| Derivation / Formula | Reasoning |
|---|---|
| \[I_{\text{sphere}} = \tfrac{2}{5}mR^{2}\] | The solid ball has a smaller inline moment of inertia, \(I\), which means it offers less resistance to changes in its rotational state. |
| \[I_{\text{cyl}} = \tfrac{1}{2}mR^{2}\] | The cylinder has a slightly larger inline moment of inertia, \(I\), and therefore resists rotational motion more than the ball. |
| \[I_{\text{sphere}} < I_{\text{cyl}}\] | Because the ball’s rotational opposition is smaller, it can start and maintain rolling motion more readily than the cylinder. |
| \[\text{Ball arrives first}\] | With lower rotational resistance, the ball reaches the bottom before the cylinder. |
| \[\text{Option A}\] | Correct: the ball’s lower \(I\) lets it roll more easily. |
| \[\text{Option B}\] | Incorrect: the cylinder’s higher \(I\) hinders its rolling compared to the ball. |
| \[\text{Option C}\] | Incorrect: identical resistance would be required for a tie, which is not the case here. |
| \[\text{Option D}\] | Incorrect: the outcome depends on shape-dependent \(I\), not mass, assuming both roll without slipping. |
| \[\text{Option E}\] | Incorrect: diameter cancels out because both shapes keep the same \(R\) in their respective expressions; the decisive factor is the coefficient in \(I\). |
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A uniform, solid, \( 100 \) \( \text{kg} \) cylinder with a diameter of \( 1.0 \) \( \text{m} \) is mounted so it is free to rotate about a fixed, horizontal, frictionless axis that passes through the centers of its circular ends. A \( 10 \) \( \text{kg} \) block is hung from a very light, thin cord wrapped around the cylinder’s circumference. When the block is released, the cord unwinds and the block accelerates downward. What is the acceleration of the block?
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?
A solid sphere \( \left( I = \frac{2}{5}MR^2 \right) \) and a solid cylinder \( \left( I = \frac{1}{2}MR^2 \right) \), both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at a point \( \frac{L}{4} \) away from the center?
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 1.00 N·m? Assume that the moment of inertia of a rectangle is \( I = \frac{1}{12}M(a^2+b^2) \)
Consider a uniform hoop of radius \( R \) and mass \( M \) rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy? Hint: The moment of inertia of a uniform hoop is \(I = M R^2\)
Which of the following must be zero if an object is spinning at a constant rate? There may be more than one right answer.
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first?
A uniform copper disk of radius \( R \) has a moment of inertia \( I \) around an axis passing through the center of the disk perpendicular to its plane. If the radius of the disk were only \( \dfrac{R}{2} \), but the thickness were the same, what would be the moment of inertia in terms of \( I \)? Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
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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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