| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(L_1 = L_2\) | Apply the conservation of angular momentum: the angular momentum at position 1 is equal to the angular momentum at position 2. |
| 2 | \(m v_1 r_1 = m v_2 r_2\) | Angular momentum (\(L\)) is defined as the product of mass (\(m\)), velocity (\(v\)), and radius (\(r\)). The mass of the comet cancels out as it is the same at both positions. |
| 3 | \(\frac{v_1}{v_2} = \frac{r_2}{r_1}\) | Rearrange the angular momentum equation to solve for the ratio of velocities. |
| 4 | \(\frac{r_1}{r_2} = 10\) | Given that the distance at position 1 (\(r_1\)) is 10 times the distance at position 2 (\(r_2\)). |
| 5 | \(\frac{v_1}{v_2} = \frac{1}{10}\) | Substitute the ratio of distances into the velocity equation to find that the speed at position 1 is one-tenth of the speed at position 2. |
| 6 | (b) \(\frac{1}{10}\) | The ratio of the speed of the comet at position 1 to its speed at position 2 is \(\frac{1}{10}\). |
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A \( 50 \, \text{kg} \) person is sitting on a seesaw \( 1.2 \, \text{m} \) from the balance point. On the other side, a \( 70 \, \text{kg} \) person is balanced. How far from the balance point is the second person sitting?
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?
A seesaw is balanced on a fulcrum, with a boy of mass \( M_1 \) sitting on one end and a girl of mass \( M_2 \) sitting on the other end. The seesaw is a uniform plank of length \( L \) and mass \( M \). The fulcrum is located at the midpoint of the plank. Does \( M_1 = M_2 \)? Justify your working.
A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom?
\( \text{Solid sphere: } I = \frac{2}{5}mR^2, \quad \text{Solid disk: } I = \frac{1}{2}mR^2, \quad \text{Hoop: } I = mR^2 \)
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 hoop with a mass \(m\) and unknown radius is rolling without slipping on a flat surface with an angular speed \(\omega\). The hoop encounters a hill and continues to roll without slipping until it reaches a maximum height \(h\).
A meter stick with a uniformly distributed mass of \(0.5 \, \text{kg}\) is supported by a pivot placed at the \(0.25 \, \text{m}\) mark from the left. At the left end, a small object of mass \(1.0 \, \text{kg}\) is placed at the zero mark, and a second small object of mass \(0.5 \, \text{kg}\) is placed at the \(0.5 \, \text{m}\) mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released.
A uniform ladder with mass \( m_2 \) and length \( L \) rests against a smooth wall. A do-it-yourself enthusiast of mass \( m_1 \) stands on the ladder a distance \( d \) from the bottom (measured along the ladder). The ladder makes an angle \( \theta \) with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude \( f \) between the floor and the ladder. \( N_1 \) is the magnitude of the normal force exerted by the wall on the ladder, and \( N_2 \) is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.
A wheel of radius \( R \) and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses \( m \), \( M \), and \( 2M \), respectively, are mounted on the rim of the wheel, as shown above. If the system is in static equilibrium, what is the value of \( m \) in terms of \( M \)?
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
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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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