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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \Delta L = \tau_{\text{net}} \times \Delta t \] | The change in angular momentum \( \Delta L \) is given by the product of the net torque \( \tau_{\text{net}} \) and the time interval \( \Delta t \). |
| 2 | \[ \tau = r \times F \times \sin(\theta) \] | The torque \( \tau \) due to a force \( F \) is the product of the force, the distance \( r \) from the axis, and the sine of the angle \( \theta \) between the force and the lever arm. |
| 3 | – | To calculate \( \tau_{\text{net}} \), you need to measure the magnitude of the forces (using a force probe) and the angles (using a protractor). |
| 4 | – | The electronic balance is not needed as mass is not required to determine the change in angular momentum. Motion sensor data is also redundant because time is already given. |
| 5 | – | Therefore, the essential measuring devices needed are a force probe and a protractor. |
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An object’s angular momentum changes by \( 10 \,\text{kg} \cdot \text{m}^2/\text{s} \) in \( 2.0 \) \( \text{s} \). What magnitude average torque acted on this object?
| Wagon | Wheel Structure | Moment of Inertia | Wheel Mass | Wheel Radius |
|---|---|---|---|---|
| Wagon \(A\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.5 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(B\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.2 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(C\) | Hollow hoop | \[M R^2\] | \[ 0.1 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
Three wagons have identical total mass (including their wheels) and each has four wheels. However, the wheels on each wagon have different designs with varying mass distributions and radii as shown in a reference chart. When accelerating each wagon from a standstill to \( 10 \) \( \text{m/s} \), which wagon requires the most energy input?
A boy is sitting at a distance [katex] d_1 [/katex] from the fulcrum, and girl is sitting at a distance [katex] d_2 [/katex] from the fulcrum, with [katex] d_1 > d_2 [/katex]. The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to ends of a rod of negligible mass of length \( 3x \). The rod is free to turn about a vertical axis through point \( P \). The first mass, \( m \), is located \( x \) away from point \( P \), and therefore the other mass, of \( 2m \), is \( 2x \) from point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_i \) about \( P \). The system is gradually brought to rest by friction.
Derive expressions for the following quantities in terms of \( \mu \), \( m \), \( x \), \( g \), and \( \omega_i \).
A centrifuge rotor rotating at \( 9200 \) \( \text{rpm} \) is shut off and is eventually brought uniformly to rest by a frictional torque of \( 1.20 \) \( \text{N} \cdot \text{m} \). If the mass of the rotor is \( 3.10 \) \( \text{kg} \) and it can be approximated as a solid cylinder of radius \( 0.0710 \) \( \text{m} \), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \( \frac{1}{2} m r^2 \).
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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] |
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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