| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[L = mvr_{\perp}\] | The angular momentum magnitude of one car about the flagpole is \(L=mvr_{\perp}\), where \(r_{\perp}\) is the perpendicular distance from the flagpole to the car’s line of motion. |
| 2 | \[v_A = \frac{d_B}{t_f}\] | Car \(A\) moves east and reaches the intersection in time \(t_f\). From the geometry, the distance Car \(A\) travels to the intersection is \(d_B\), so its constant speed is \(v_A=\frac{d_B}{t_f}\). |
| 3 | \[v_B = \frac{d_A}{t_f}\] | Car \(B\) moves north and reaches the intersection in time \(t_f\). From the geometry, the distance Car \(B\) travels to the intersection is \(d_A\), so its constant speed is \(v_B=\frac{d_A}{t_f}\). |
| 4 | \[L_A = m_A v_A d_A\] | For Car \(A\), the perpendicular distance from the flagpole to its road is \(d_A\). Therefore, its angular momentum magnitude about the flagpole is \(L_A=m_Av_Ad_A\). |
| 5 | \[L_A = m_A\left(\frac{d_B}{t_f}\right)d_A\] | Substitute \(v_A=\frac{d_B}{t_f}\) into the expression for Car \(A\)’s angular momentum. |
| 6 | \[L_A = \frac{m_A d_A d_B}{t_f}\] | This is the magnitude of Car \(A\)’s angular momentum about the flagpole at \(t=0\). |
| 7 | \[L_B = m_B v_B d_B\] | For Car \(B\), the perpendicular distance from the flagpole to its road is \(d_B\). Therefore, its angular momentum magnitude about the flagpole is \(L_B=m_Bv_Bd_B\). |
| 8 | \[L_B = m_B\left(\frac{d_A}{t_f}\right)d_B\] | Substitute \(v_B=\frac{d_A}{t_f}\) into the expression for Car \(B\)’s angular momentum. |
| 9 | \[L_B = \frac{m_B d_A d_B}{t_f}\] | This is the magnitude of Car \(B\)’s angular momentum about the flagpole at \(t=0\). |
| 10 | \[L_{\text{total}} = \left|L_B – L_A\right|\] | The angular momenta point in opposite directions because one car’s motion produces clockwise rotation about the flagpole and the other produces counterclockwise rotation. Therefore, the total angular momentum magnitude is the absolute value of the difference. |
| 11 | \[L_{\text{total}} = \left|\frac{m_B d_A d_B}{t_f} – \frac{m_A d_A d_B}{t_f}\right|\] | Substitute the expressions for \(L_A\) and \(L_B\) into the total angular momentum expression. |
| 12 | \[L_{\text{total}} = \left|\frac{(m_B-m_A)d_A d_B}{t_f}\right|\] | Factor out the common term \(\frac{d_A d_B}{t_f}\). Among the choices, this corresponds to choice \(\text{B}\), assuming the listed expression represents the signed difference whose magnitude is intended. |
| 13 | \[\boxed{\frac{(m_B-m_A)d_A d_B}{t_f}}\] | The correct multiple-choice answer is \(\text{B}\). Choice \(\text{A}\) would be correct only if the two angular momenta were in the same direction. Choices \(\text{C}\) and \(\text{D}\) include only one car’s contribution, so each ignores the angular momentum of the other car. |
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A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
A man with mass \( m \) is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius \( R \) that rotates without friction at a constant angular velocity \( \omega \). Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.
Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since \( L \propto I \), the angular momentum will decrease proportionately and the platform will slow down.
Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.
Explain what is correct and incorrect about each students statement if anything.
A disk of known radius and rotational inertia can rotate without friction in a horizontal plane around its fixed central axis. The disk has a cord of negligible mass wrapped around its edge. The disk is initially at rest, and the cord can be pulled to make the disk rotate. Which of the following procedures would best determine the relationship between applied torque and the resulting change in angular momentum of the disk?
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(2M\) that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?
The elliptical orbit of a comet is shown above. Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2. At position 2, the comet’s kinetic energy is
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to a rod of negligible mass of length \( 3l \) as shown above. The rod is free to turn about a vertical axis through point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_0 \) about \( P \). The system is gradually brought to rest by friction. Develop expressions for the following quantities in terms of \( \mu \), \( m \), \( l \), \( g \), and \( \omega_0 \).
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(5M\)that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will result in an increase in the total angular momentum of the child-platform system?
Angular momentum cannot be conserved if
A spinning ice skater on extremely smooth ice is able to control the rate at which she rotates by pulling in her arms. Which of the following statements are true about the skater during this process?
During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which was used to create the graph that is shown. A frictional torque is exerted on the wheel. A student makes the following statement about the data. “The frictional torque exerted on the wheel is independent of the wheel’s angular speed.” Does the data from the graph support the student’s statement? Justify your selection.
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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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