| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) | \[\alpha = \frac{\Delta\omega}{\Delta t}=\frac{-30-60}{6-2}=\frac{-90}{4}=-22.5\,\text{rad\,/s}^2\] | The interval from \(t=2\,\text{s}\) to \(t=6\,\text{s}\) is a straight line, so the slope (constant) gives the instantaneous angular acceleration at any time in that portion, including \(t=4\,\text{s}.\) |
| (b) | \[v=r\,\omega = 0.25\,\text{m}\times 60\,\text{rad\,/s}=15\,\text{m\,/s}\] | At \(t=1\,\text{s}\) the graph is flat at \(\omega = 60\,\text{rad\,/s}.\) Linear (rim) speed is \(v=r\omega.\) |
| (c) | \[\Delta\theta_{0\!\to 2}=\omega\,\Delta t = 60\,\text{rad\,/s}\times 2\,\text{s}=120\,\text{rad}\] | From 0–2 s the angular velocity is constant, so the area under the \(\omega\)-vs-\(t\) graph (a rectangle) is \(\omega\Delta t.\) |
| (d) | \[\omega_f = \omega_i + \alpha\,\Delta t = 60 + (-22.5)(2)=15\,\text{rad\,/s}\] \[\Delta\theta_{2\!\to 4}=\tfrac12(\omega_i+\omega_f)\,\Delta t = \tfrac12(60+15)\times 2 = 75\,\text{rad}\] |
The segment 2–4 s lies on the linear portion with constant \(\alpha.\) Use kinematics (or trapezoid area) with \(\omega_i=60\,\text{rad\,/s}\) and \(\omega_f=15\,\text{rad\,/s}.\) |
| (e) | \[\begin{aligned} \Delta\theta_{0\!\to 2}&=120\\[4pt] \Delta\theta_{2\!\to 6}&=\tfrac12(60+(-30))(4)=60\\[4pt] \Delta\theta_{6\!\to 8}&=(-30)(2)=-60\\[4pt] \Delta\theta_{8\!\to 10}&=\tfrac12(-30+0)(2)=-30\\[4pt] \theta_{\text{total}}&=120+60-60-30=90\,\text{rad} \end{aligned}\] |
Sum the signed areas (trapezoids/rectangles) for each time interval. Positive areas correspond to counter-clockwise rotation; negative areas to clockwise. |
| (f) | \[\Delta x = r\,\theta_{\text{total}} = 0.25\,\text{m}\times 90\,\text{rad}=22.5\,\text{m}\] | For rolling without slipping, the center of mass translates a linear distance equal to \(r\,\Delta\theta.\) |
| (g) | \[a_{\text{tan}} = r\,|\alpha| = 0.25\,\text{m}\times 22.5\,\text{rad\,/s}^2 = 5.6\,\text{m\,/s}^2\] | Magnitude of tangential acceleration is the product of radius and magnitude of angular acceleration at \(t=4\,\text{s}.\) |
| (h) | \[a_{\text{tan}} = r\,\alpha = 0.25\times 0 = 0\,\text{m\,/s}^2\] \[v = r\omega = 0.25\times 60 = 15\,\text{m\,/s}\quad\Rightarrow\quad a_{c}=\frac{v^2}{r}=\frac{15^2}{0.25}=900\,\text{m\,/s}^2\] |
At \(t=1\,\text{s}\) the graph is flat, so \(\alpha=0\Rightarrow a_{\text{tan}}=0.\) Centripetal acceleration depends on instantaneous speed: \(a_c = v^2/r.\) |
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A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously from the top of an inclined plane and do not slip, which one will reach the bottom first?

The rotating systems, shown in the figure above, differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left), or a distance r/2 from the axis of rotation (right). What happens if you release the hanging blocks simultaneously from rest?
Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at \( 30^\circ \) to the plane of the door. Which force exerts the greater torque about the door hinge?

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?
In a demonstration, a teacher holds the axle of a wheel that is spinning with constant angular speed. The teacher then releases the axle and the wheel begins to fall toward the ground. As the wheel falls, its angular speed remains constant. Which of the following correctly describes how the rotational kinetic energy \( K_{\text{rot}} \) of the wheel and the total kinetic energy \( K_{\text{tot}} \) of the wheel change, if at all, after the wheel is released but before it reaches the ground?
| \( K_{\text{rot}} \) | \( K_{\text{tot}} \) | |
|---|---|---|
| A | Constant | Constant |
| B | Constant | Increasing |
| C | Increasing | Constant |
| D | Increasing | Increasing |
A student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is \(I = MR^2\). Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers.
A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal, \( 2.3 \) \( \text{m} \)-long, \( 6.1 \) \( \text{kg} \) reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A \( 64 \) \( \text{kg} \) woman lies on the reaction board with her feet over the pivot. The scale reads \( 27 \) \( \text{kg} \). What is the distance from the woman’s feet to her center of mass? Express your answer with the appropriate units.
A force of \(17 \, \text{N}\) is applied to the end of a \(0.63 \, \text{m}\) long torque wrench at an angle \(45^\circ\) from a line joining the pivot point to the handle. What is the magnitude of the torque about the pivot point produced by this force?
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 \( 80.0 \) \( \text{rad/s}^2 \).
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).
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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