| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 (a) | \( v = r \omega \) | For rolling motion, the linear velocity \( v \) at the bottom is related to the angular velocity \( \omega \) by the radius \( r \). Here, the radius \( r \) is 0.16 m. |
| 2 (a) | \( \omega = \frac{v}{r} \) | Calculate the angular velocity \( \omega \) at the bottom using the given linear velocity \( v = 3.2 \, \text{m/s} \). |
| 3 (a) | \( \omega = \frac{3.2 \, \text{m/s}}{0.16 \, \text{m}} = 20 \, \text{rad/s} \) | Substitute the values into the equation to find \( \omega \). |
| 4 (a) | \( \alpha = \frac{\omega}{t} \) | Angular acceleration \( \alpha \) is calculated using the angular velocity \( \omega \) and the time \( t \) it takes to reach that angular velocity. |
| 5 (a) | \( v^2 = v_0^2 + 2aL \) | Using the kinematic equation for linear motion, where \( a \) is linear acceleration and \( L = 1.5 \, \text{m} \) is the length of the incline. |
| 6 (a) | \( 3.2^2 = 0 + 2a \times 1.5 \) | \( a = \frac{(3.2)^2}{2 \times 1.5} = \frac{10.24}{3} \approx 3.413 \, \text{m/s}^2 \) |
| 7 (a) | \( a = r\alpha \) | Relate linear acceleration \( a \) to angular acceleration \( \alpha \). |
| 8 (a) | \( \alpha = \frac{a}{r} = \frac{3.413 \, \text{m/s}^2}{0.16 \, \text{m}} \approx 21.33 \, \text{rad/s}^2 \) | Compute angular acceleration \( \alpha \). |
| 1 (b) | \( \omega = \alpha t \) | Use the equation for angular velocity \( \omega \) related to angular acceleration \( \alpha \) and time \( t \). |
| 2 (b) | \( \omega_{\text{rpm}} = 7329 = \omega \frac{60}{2\pi} \) | Convert \( \omega \) from rpm to rad/s for calculation. |
| 3 (b) | \( \omega = 7329 \cdot \frac{2\pi}{60} \approx 767.23 \, \text{rad/s} \) | Find \( \omega \) in rad/s. |
| 4 (b) | \( t = \frac{\omega}{\alpha} = \frac{767.23}{419} \approx 1.83 \, \text{s} \) | Compute the time \( t \). |
| 1 (c) | \( \Delta \omega = \alpha \Delta t \) | Angular velocity change \( \Delta \omega \) is given, calculate the time \( \Delta t \) using angular acceleration \( \alpha \). |
| 2 (c) | \( \Delta \omega = 33.3 \, \text{rad/s} – 3.33 \, \text{rad/s} = 29.97 \, \text{rad/s} \) | Calculate the change in angular velocity \( \Delta \omega \). |
| 3 (c) | \( \Delta t = \frac{\Delta \omega}{\alpha} = \frac{29.97}{5.15} \approx 5.82 \, \text{s} \) | Calculate the time \( \Delta t \) needed to reach the final angular velocity. |
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A uniform stick has length \( L \). The moment of inertia about the center of the stick is \( I_0 \). A particle of mass \( M \) is attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
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
The angular velocity of a rotating disk of radius \(20 \, \text{cm}\) increases from \(1 \, \text{rad/s}\) to \(3 \, \text{rad/s}\) in \(0.5 \, \text{s}\). What is the linear tangential acceleration of a point on the rim of the disk during this time interval?
A uniform solid cylinder of mass \( M \) and radius \( R \) is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
In lacrosse, a typical throw is made by rotating the stick through an angle of roughly \(90^\circ\), then releasing the ball when the stick is vertical, as shown above. If the \(1 \, \text{meter}\) long stick is at rest when horizontal and the ball leaves the stick with a velocity of \(10 \, \text{m/s}\), what angular acceleration must the stick experience?
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?
Two masses, \( m_y = 32 \) \( \text{kg} \) and \( m_z = 38 \) \( \text{kg} \), are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius \( R = 0.311 \) \( \text{m} \) and mass \( 3.1 \) \( \text{kg} \). Initially, \( m_y \) is on the ground and \( m_z \) rests \( 2.5 \) \( \text{m} \) above the ground.
A uniform rod of length \( L \) is pivoted at one end \(45^{\circ}\) below the horizontal and released from rest. The rod swings freely downward. Which of the following best describes the angular acceleration of the rod as it swings from the initial position to the vertical position?
A light string is attached to a massive pulley of known rotational inertia \( I_P \), as shown in the figure. A student must determine the relationship between the torque exerted on the pulley and the change in the pulley’s angular velocity when the torque is applied for \( 2.0 \) \( \text{s} \). In addition to a stopwatch to measure the time interval, what two measurements could the student make in order to determine the relationship? Select two answers.
(a) Angular acceleration of the ball is \( \approx 21.33 \, \text{rad/s}^2 \).
(b) Time for the CD player to reach full speed is \( \approx 1.83 \, \text{s} \).
(c) Time to accelerate for the rotating object is \( \approx 5.82 \, \text{s} \).
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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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