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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[r = \frac{d}{2} = \frac{20\,\text{cm}}{2} = 0.10\,\text{m}\] | The radius is half of the \(20\,\text{cm}\) diameter. |
| 2 | \[\omega = 20000\,\text{rpm} \times \frac{2\pi\,\text{rad}}{1\,\text{rev}} \times \frac{1\,\text{min}}{60\,\text{s}}\] | Convert \(\omega\) from revolutions per minute to radians per second. |
| 3 | \[\omega = 2.094\times10^{3}\,\text{rad/s}\] | Simplified numerical value of \(\omega\). |
| 4 | \[v_x = \omega r\] | Linear speed on the rim relates to angular speed by \(v_x = \omega r\). |
| 5 | \[\boxed{v_x \approx 2.09\times10^{2}\,\text{m/s}}\] | Multiplying \(\omega\) and \(r\) gives the rim speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\omega_i = 2.094\times10^{3}\,\text{rad/s}\qquad \omega_f = 0.60\,\omega_i\] | Initial angular velocity \(\omega_i\) from part (a); final is \(40\%\) lower (i.e., \(60\%\) of \(\omega_i\)). |
| 2 | \[\alpha = \frac{\omega_f-\omega_i}{\Delta t}\] | For constant angular acceleration, \(\alpha = \Delta\omega/\Delta t\). |
| 3 | \[\alpha = \frac{-0.40\,\omega_i}{30\,\text{s}} = -2.79\times10^{1}\,\text{rad/s}^2\] | Substitute \(\omega_i\) and \(\Delta t = 30\,\text{s}\). The negative sign shows the wheel is slowing. |
| 4 | \[\boxed{|\alpha| \approx 2.79\times10^{1}\,\text{rad/s}^2}\] | Report the magnitude of the angular acceleration. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\theta = \frac{\omega_i + \omega_f}{2}\,\Delta t\] | With constant \(\alpha\), average angular speed is \((\omega_i+\omega_f)/2\). |
| 2 | \[\theta = \frac{2.094\times10^{3}+1.256\times10^{3}}{2}\times30\,\text{s}\] | Insert \(\omega_i\) and \(\omega_f\). |
| 3 | \[\theta \approx 5.03\times10^{4}\,\text{rad}\] | Calculate total angular displacement in radians. |
| 4 | \[N = \frac{\theta}{2\pi} \approx 8.0\times10^{3}\,\text{rev}\] | Divide by \(2\pi\) to convert radians to revolutions. |
| 5 | \[\boxed{N \approx 8.0\times10^{3}\,\text{rev}}\] | Number of complete revolutions in 30 s. |
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Four systems are in rotational motion. Which of the following combinations of rotational inertia and angular speed for each of the systems corresponds to the greatest rotational kinetic energy?
| System | Rotational Inertia | Angular Speed |
|---|---|---|
| A | \( I_0 \) | \( \omega_0 \) |
| B | \( I_0 \) | \( 4\, \omega_0 \) |
| C | \( 2 I_0 \) | \( 2\, \omega_0 \) |
| D | \( 6 I_0 \) | \( \omega_0 \) |
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.
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.
A traffic light hangs from a pole as shown in the diagram. The uniform aluminum pole AB is of length \( 7.20 \) \( \text{m} \) and has a mass of \( 12.0 \) \( \text{kg} \). The mass of the traffic light is \( 21.5 \) \( \text{kg} \). The point C is located \( 3.80 \) \( \text{m} \) vertically above the pivot A. A massless horizontal cable CD is attached at C and connects to the pole at point D, where the pole makes an angle of \( 37^{\circ} \) with the cable.
What is the rotational inertia \( I \) of a disk with a radius \( R = 4 \) \( \text{m} \) and a mass \( 2 \) \( \text{kg} \)? The same disk is rotated around an axis that is \( 0.5 \) \( \text{m} \) from the center of the disk. What is the new rotational inertia \( I \) of the disk? What would the rotational inertia be if the disk axis was \( 3.75 \) \( \text{m} \) from the center?
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?
The system in the Figure is in equilibrium. A concrete block of mass 225 kg hangs from the end of a uniform strut whose mass is 45.0 kg.
The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?
An isolated spherical star of radius \( R_o \), rotates about an axis that passes through its center with an angular velocity of \( \omega_o \). Gravitational forces within the star cause the star’s radius to collapse and decrease to a value \( r_o < R_o \), but the mass of the star remains constant. A graph of the star’s angular velocity as a function of time as it collapses is shown. Which of the following predictions is correct about the angular momentum \( L \) of the star immediately after the collapse?
\(2.09\times10^{2}\,\text{m/s}\)
\(2.79\times10^{1}\,\text{rad/s^{2}}\)
\(8.0\times10^{3}\,\text{rev}\)
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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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