| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[I = \tfrac{1}{2} M R^{2}\] | Moment of inertia of a uniform solid disk; here \(M = 2.00\,\text{kg}\) and \(R = 0.0700\,\text{m}\). |
| 2 | \[I = 0.00490\,\text{kg\,m}^2\] | Substituting \(M\) and \(R\) into the formula: \(0.5\times2.00\times0.0700^{2}\). |
| 3 | \[\alpha = \frac{\tau}{I}\] | Constant angular acceleration produced by the motor torque \(\tau = 0.600\,\text{N m}\). |
| 4 | \[\alpha = 1.22\times10^{2}\,\text{rad/s}^2\] | Numeric evaluation: \(\alpha = 0.600 / 0.00490\). |
| 5 | \[\omega_f = 1200\, \text{rev/min} = 40\pi\,\text{rad/s}\] | Convert final speed using \(2\pi\,\text{rad}=1\,\text{rev}\) and \(60\,\text{s}=1\,\text{min}\). |
| 6 | \[t = \frac{\omega_f}{\alpha}\] | For uniform acceleration from rest, \(\omega_f = \alpha t\). |
| 7 | \[t = 1.03\,\text{s}\] | Compute time with \(\omega_f = 40\pi\,\text{rad/s}\) and \(\alpha = 1.22\times10^{2}\,\text{rad/s}^2\). |
| 8 | \[\boxed{t = 1.03\,\text{s}}\] | Time required to reach operating speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\theta = \tfrac{1}{2} \alpha t^{2}\] | Angular displacement under constant \(\alpha\) starting from rest. |
| 2 | \[\theta = 6.47\times10^{1}\,\text{rad}\] | Substitute \(\alpha = 1.22\times10^{2}\,\text{rad/s}^2\) and \(t = 1.03\,\text{s}\): \(0.5\times122\times1.03^{2}=64.7\,\text{rad}\). |
| 3 | \[N = \frac{\theta}{2\pi}\] | Convert radians to revolutions; one revolution is \(2\pi\,\text{rad}\). |
| 4 | \[N = 1.03\times10^{1}\,\text{rev}\] | Numeric result: \(64.7 / 6.283 = 10.3\,\text{rev}\). |
| 5 | \[\boxed{N \approx 10.3\,\text{rev}}\] | Total revolutions made while accelerating. |
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A uniform rod of mass \( M_0 \) and length \( L \) is free to rotate about a pivot at its left end and is released from rest when the rod is \( 30^{\circ} \) below the horizontal, as shown in the figure. With respect to the pivot, the rod has rotational inertia \( I_0 = \dfrac{1}{3} M_0 L^2 \). Which of the following expressions correctly represents the magnitude of the net torque exerted on the rod about the pivot at the moment the rod is released?
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).
A car accelerates from \( 0 \) to \( 25 \) \( \text{m/s} \) in \( 5 \) \( \text{s} \). If the car’s tires have a diameter of \( 70 \) \( \text{cm} \), how many revolutions does a tire make while accelerating?
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
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.
Initially, a ball has an angular velocity of \( 5.0 \) \( \text{rad/s} \) counterclockwise. Some time later, after rotating through a total angle of \( 5.5 \) \( \text{radians} \), the ball has an angular velocity of \( 1.5 \) \( \text{rad/s} \) clockwise.
A disk increases from 2 complete revolutions in 2 seconds to 5 complete revolutions in 2 seconds. What is its average angular acceleration?
A miniature, solid globe with mass \( 0.25 \) \( \text{kg} \) and radius \( 0.10 \) \( \text{m} \) is spinning in place about a vertical axis with the equator horizontal, as shown. A point on the globe’s equator, represented by the dot in the figure, has a linear speed of \( 4.0 \) \( \text{m/s} \). The rotational inertia of a solid sphere of mass \( m \) and radius \( r \) is \( \tfrac{2}{5}mr^{2} \). The rotational kinetic energy of the globe is most nearly
A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?
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.
\(1.03\,\text{s}\)
\(10.3\,\text{rev}\)
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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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