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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\alpha = 2\pi \, \text{rad/s}^2\) | Given angular acceleration. |
| 2 | \(r = 2 \, \text{m}\) | Given radius of the wheel. |
| 3 | \(t = 3 \, \text{s}\) | Given time period over which the wheel accelerates. |
| 4 | \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\) | Formula for angular displacement. Here, \( \omega_i \) is the initial angular velocity which is zero because the wheel is initially at rest. |
| 5 | \(\theta = 0 \cdot t + \frac{1}{2} \cdot 2\pi \cdot (3)^2\) | Substituting the given values into the formula. Since \( \omega_i = 0 \), the first term vanishes. |
| 6 | \(\theta = \frac{1}{2} \cdot 2\pi \cdot 9\) | Simplifying the expression for angular displacement. |
| 7 | \(\theta = 9\pi \, \text{rad}\) | Calculate the angular displacement after 3 seconds. |
| 8 | \(\Delta x = r \theta\) | Formula for linear displacement, utilizing the relationship between linear and angular displacement. |
| 9 | \(\Delta x = 2 \cdot 9\pi\) | Substituting the values of \( r \) and \( \theta \) to find the linear displacement. |
| 10 | \(\Delta x = 18\pi \, \text{m}\) | Final calculation of the distance traveled by the wheel after 3 seconds. |
| 11 | Option (d) | The correct answer is \( 18\pi \, \text{m} \). This corresponds to choice (d). |
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A rotating merry-go-round makes one complete revolution in 4.0 s. What is the linear speed and acceleration of a child seated 1.2 m from the center?
In both cases, a massless rod is supported by a fulcrum, and a \(200 \, \text{kg}\) hanging mass is suspended from the left end of the rod by a cable. A downward force \(F\) keeps the rod in rest. The rod in Case A is \(50 \, \text{cm}\) long, and the rod in Case B is \(40 \, \text{cm}\) long (each rod is marked at \(10 \, \text{cm}\) intervals). The magnitude of each vertical force \(F\) exerted on the rod will be
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? [katex] I_{sphere} = \frac{2}{5}MR^2[/katex], [katex] I_{cylinder} = \frac{1}{2}MR^2[/katex], [katex] I_{pipe} = MR^2[/katex]
A windmill blade with a rotational inertia of \( 6.0 \) \( \text{kg} \cdot \text{m}^2 \) has an initial angular velocity of \( 8 \) \( \text{rad/s} \) in the clockwise direction. It is then given an angular acceleration of \( 4 \) \( \text{rad/s}^2 \) in the clockwise direction for \( 10 \) seconds. What is the change in rotational kinetic energy of the blade over this time interval?
Why are doorknobs located on the side of the door opposite the hinges?
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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] |
General Metric Conversion Chart
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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