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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$ T_{1} – m_{1}g = m_{1}a $$ | This is Newton’s second law for mass \(m_{1}\) moving upward. |
| 2 | $$ m_{2}g – T_{2} = m_{2}a $$ | This is Newton’s second law for mass \(m_{2}\) moving downward. |
| 3 | $$ T_{1} = m_{1}g + m_{1}a \quad \text{and} \quad T_{2} = m_{2}g – m_{2}a $$ | Rearrange the equations to solve for the tensions in the string. |
| 4 | $$ (T_{2} – T_{1})R = I\left(\frac{a}{R}\right) $$ | This relates the net torque on the pulley to its moment of inertia \(I\) using the no‐slip condition \(\alpha = \frac{a}{R}\). |
| 5 | $$ T_{2} – T_{1} = \frac{I\,a}{R^{2}} $$ | Simplify the torque equation by dividing both sides by \(R\). |
| 6 | $$ (m_{2}g – m_{2}a) – (m_{1}g + m_{1}a) = $$$$(m_{2}-m_{1})g – (m_{2}+m_{1})a =$$$$\frac{I\,a}{R^{2}} $$ | Substitute the expressions for \(T_{1}\) and \(T_{2}\) into the torque equation. |
| 7 | $$ I = \frac{R^{2}}{a}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})a\Bigr] $$ | Rearrange the equation to solve for the moment of inertia \(I\). |
| 8 | $$ h = \frac{1}{2}at^{2} $$ | Use the kinematics relation for the heavy mass \(m_{2}\) falling a distance \(h\) from rest. |
| 9 | $$ a = \frac{2h}{t^{2}} $$ | Solve for the acceleration \(a\) from the kinematics equation. |
| 10 | $$ I = \frac{R^{2}}{\frac{2h}{t^{2}}}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})\frac{2h}{t^{2}}\Bigr] $$ | Substitute \(a = \frac{2h}{t^{2}}\) into the expression for \(I\). |
| 11 | $$ I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1}) $$ | Simplify the expression to obtain \(I\) solely in terms of \(m_{1}, m_{2}, R, h, t\) and \(g\). |
| 12 | $$ \boxed{I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})} $$ | This is the final algebraic expression for the pulley’s moment of inertia. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$ \Delta x = R\theta $$ | This equation relates the linear displacement \(\Delta x\) to the angular displacement \(\theta\) of the pulley. |
| 2 | $$ h = R\theta $$ | Since the heavy mass \(m_{2}\) falls a distance \(h\), the length of the unwound rope is \(h\), which equals \(R\theta\). |
| 3 | $$ \theta = \frac{h}{R} $$ | Solve for the angular displacement \(\theta\) of the pulley. |
| 4 | $$ \boxed{\theta = \frac{h}{R}} $$ | This is the final expression for the total rotation of the pulley in radians. |
Just ask: "Help me solve this problem."
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 [katex] I = \frac{1}{12}M(a^2+b^2) [/katex]
A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting?
Consider a solid uniform sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?
An airliner arrives at the terminal, and the engines are shut off. The rotor of one of the engines has an initial clockwise angular velocity of \( 2000 \) \( \text{rad/s} \). The engine’s rotation slows with an angular acceleration of magnitude \( 80.0 \) \( \text{rad/s}^2 \).
Two spheres of equal size and equal mass are rotated with an equal amount of torque. One of the spheres is solid with its mass evenly distributed throughout its volume, and the other is hollow with all of its mass concentrated at the edges. Which sphere would rotate faster?
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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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