| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\omega_x = \omega_i + \alpha t\] | Rotational kinematic relation where \(\omega_i\) is the initial angular velocity, \(\alpha\) the (constant) angular acceleration, and \(t\) the elapsed time. |
| 2 | \[\omega_x = 2000\,\text{rad/s} – 80.0\,\text{rad/s}^2(10.0\,\text{s})\] | The engine is slowing, so the angular acceleration is opposite the initial clockwise motion, giving \(\alpha = -80.0\,\text{rad/s}^2\). |
| 3 | \[\boxed{\omega_x = 1200\,\text{rad/s (clockwise)}}\] | Simplifying the expression yields the angular velocity after \(10\,\text{s}\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[0 = \omega_i + \alpha t\] | The rotor comes to rest when \(\omega_x = 0\). |
| 2 | \[t = \frac{-\omega_i}{\alpha} = \frac{-2000\,\text{rad/s}}{-80.0\,\text{rad/s}^2}\] | Rearrange the kinematic equation to solve for \(t\). |
| 3 | \[\boxed{t = 25\,\text{s}}\] | Compute the time required for the rotor to stop completely. |
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Two points, A and B, are on a disk that rotates about an axis. Point A is \( 3 \) times as far from the axis as point B. If the speed of point B is \( v \), then what is the speed of point A?
A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of \( 2\pi \, \text{rad/s}^2 \). If the wheel has a radius of \( 2 \, \text{m} \), what distance does the wheel travel in \( 3 \) seconds?
A motorcycle has tires with a diameter of \( 44.0 \) \( \text{cm} \). Cruising down the highway, they are rotating at \( 1150 \) \( \text{rpm} \) (revolutions per minute).
At time \( t = 0 \), a disk starts from rest and begins spinning about its center with a constant angular acceleration of magnitude \( \alpha \). At time \( t_f \), the disk has angular speed \( \omega_f \). Which of the following expressions correctly compares the final angular displacement \( \theta_f \) of the disk at time \( t_f \) to the angular displacement \( \theta_{1/2} \) at time \( \frac{t_f}{2} \)?
A solid sphere \( I = 0.06 \, \text{kg} \cdot \text{m}^2 \) spins freely around an axis through its center at an angular speed of \( 20 \, \text{rad/s} \). It is desired to bring the sphere to rest by applying a friction force of magnitude \( 2.0 \, \text{N} \) to the sphere’s outer surface, a distance of \( 0.30 \, \text{m} \) from the sphere’s center. How much time will it take the sphere to come to rest?
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 \( 80.0 \) \( \text{rad/s}^2 \).
The driver of a car traveling at \( 30.0 \) \( \text{m/s} \) applies the brakes and undergoes a constant negative acceleration of \( 2.00 \) \( \text{m/s}^2 \). How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of \( 0.300 \) \( \text{m} \)?
Young David experimented with slings before tackling Goliath. He found that he could develop an angular speed of \( 8.0 \) \( \text{rev/s} \) in a sling \( 0.60 \) \( \text{m} \) long. If he increased the length to \( 0.90 \) \( \text{m} \), he could revolve the sling only \( 6.0 \) times per second.
A wheel 31 cm in diameter accelerates uniformly from 240rpm to 360rpm in 6.8 s. How far will a point on the edge of the wheel have traveled in this time?
The angular velocity of an electric motor is \(\omega = \left(20 – \frac{1}{2} t^2 \right) \, \text{rad/s}\), where \(t\) is in seconds.
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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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