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| Derivation or Formula | Reasoning |
|---|---|
| \[\sum \tau_{\text{about fulcrum}} = 0\] | For the seesaw to be balanced (no rotation), the net torque about the fulcrum must be zero. |
| \[r_1 = \frac{L}{2},\quad r_2 = \frac{L}{2}\] | Each child sits at an end of the plank, and the fulcrum is at the midpoint, so each is the same distance \( \frac{L}{2} \) from the pivot. |
| \[W_1 = M_1 g,\quad W_2 = M_2 g,\quad W_{\text{plank}} = Mg\] | The downward forces are the weights of the boy, the girl, and the uniform plank, each given by \(mg\). |
| \[r_{\text{plank}} = 0\] | The plank is uniform, so its center of mass is at its midpoint, which coincides with the fulcrum; thus its lever arm about the fulcrum is \(0\). |
| \[\tau_1 = M_1 g\left(\frac{L}{2}\right)\] | The boy’s weight produces a torque of magnitude \(rF = \left(\frac{L}{2}\right)(M_1 g)\) about the fulcrum. |
| \[\tau_2 = M_2 g\left(\frac{L}{2}\right)\] | The girl’s weight produces a torque of magnitude \(rF = \left(\frac{L}{2}\right)(M_2 g)\) about the fulcrum on the opposite side. |
| \[\tau_{\text{plank}} = Mg(0) = 0\] | The plank’s weight acts at the fulcrum, so it produces no torque and does not affect the balancing condition. |
| \[M_1 g\left(\frac{L}{2}\right) – M_2 g\left(\frac{L}{2}\right) = 0\] | Choose a sign convention (e.g., boy’s side torque positive, girl’s side negative). Balance requires the torques cancel. |
| \[g\left(\frac{L}{2}\right)(M_1 – M_2)=0\] | Factor out the common nonzero constants \(g\) and \(\frac{L}{2}\). |
| \[M_1 – M_2 = 0 \quad \Rightarrow \quad \boxed{M_1 = M_2}\] | Since \(g \neq 0\) and \(\frac{L}{2} \neq 0\), the only way for the product to be zero is \(M_1=M_2\). Therefore, for a balanced seesaw with equal distances from the fulcrum, the masses must be equal. |
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An object is experiencing a nonzero net force. Which of the following statements is most accurate?
The object shown in the diagram below consists of a cylinder of mass \( 100 \) \( \text{kg} \) and radius \( 25.0 \) \( \text{cm} \) connected by four thin rods, each of mass \( 5.00 \) \( \text{kg} \) and length \( 0.75 \) \( \text{m} \), to a thin-outer ring of mass \( 20.0 \) \( \text{kg} \). A small chunk of metal of mass \( 1.00 \) \( \text{kg} \) is welded to the outer ring. Determine the moment of inertia of the entire assembly about the center of the inner cylinder, treating the metal chunk as a point mass. Hint: The moment of inertia of a disk about it center is \(\tfrac{1}{2} M R^2\), a thin rod about it center is \(\tfrac{1}{12}ML^2\), and a thin hoop about its center is \(I = MR^2\).
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 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.
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
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
A uniform meter stick has a mass of \( 45.0 \) \( \text{g} \) placed at the \( 20 \) \( \text{cm} \) mark. If a pivot is placed at the \( 42.5 \) \( \text{cm} \) mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick?
A rod may freely rotate about an axis that is perpendicular to the rod and is along the plane of the page. The rod is divided into four sections of equal length of 0.2 m each, and four forces are exerted on the rod, as shown in the figure. Frictional forces are considered negligible. Which of the following describes an additional torque that must be applied in order to keep the rod from rotating?
A \( 0.72 \) \( \text{m} \)-diameter solid sphere can be rotated about an axis through its center by a torque of \( 10.8 \) \( \text{Nm} \) which accelerates it uniformly from rest through a total of \( 160 \) revolutions in \( 15.0 \) \( \text{s} \). What is the mass of the sphere?
Three forces of equal magnitude are applied to a \( 3 \)-m by \( 2 \)-m rectangle. Force \( F_1 \) and \( F_2 \) act at \( 45^\circ \) angles to the vertical as shown, while \( F_3 \) acts horizontally.
Yes. Both children have identical masses. See working in explanation.
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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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