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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | $$v = r\omega$$ | Linear speed \(v\) in uniform circular motion relates to angular velocity \(\omega\) through radius \(r\). |
| 2 | $$\omega = \frac{v}{r}$$ | Rearrange to solve for \(\omega\). |
| 3 | $$\omega = \frac{80\,\text{m/s}}{200\,\text{m}} = 0.4\,\text{rad/s}$$ | Substitute given \(v\) and \(r\) to obtain the numerical value. |
| 4 | $$\boxed{0.4\,\text{rad/s}}$$ | Final answer for angular velocity. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | $$a_t = \frac{dv}{dt}$$ | Tangential acceleration \(a_t\) is the time rate of change of the linear speed. |
| 2 | $$a_t = 0$$ | The car moves with constant speed, so \(\frac{dv}{dt}=0\). |
| 3 | $$\boxed{0\,\text{m/s}^2}$$ | Final answer for tangential acceleration. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | $$\alpha = \frac{d\omega}{dt}$$ | Rotational (angular) acceleration \(\alpha\) is the time rate of change of angular velocity. |
| 2 | $$\alpha = 0$$ | Since \(\omega\) is constant (from part a), \(\frac{d\omega}{dt}=0\). |
| 3 | $$\boxed{0\,\text{rad/s}^2}$$ | Final answer for rotational acceleration. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | $$a_c = \frac{v^2}{r}$$ | Centripetal acceleration \(a_c\) points toward the center during uniform circular motion. |
| 2 | $$a_c = \frac{80^2}{200} = 32\,\text{m/s}^2$$ | Insert \(v = 80\,\text{m/s}\) and \(r = 200\,\text{m}\). |
| 3 | $$\boxed{32\,\text{m/s}^2}$$ | Final answer for centripetal acceleration. |
Just ask: "Help me solve this problem."
A solid sphere of mass [katex] 1.5 \, \text{kg} [/katex] and radius [katex] 15 \, \text{cm} [/katex] rolls without slipping down a [katex] 35^\circ[/katex] incline that is [katex] 7 \, \text{m} [/katex] long. Assume it started from rest. The moment of inertia of a sphere is [katex] I= \frac{2}{5}MR^2 [/katex].
A point on the edge of a disk rotates around the center of the disk with an initial angular velocity of 3 rad/s clockwise. The graph shows the point’s angular acceleration as a function of time. The positive direction is considered to be counterclockwise. All frictional forces are considered to be negligible.
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?
When a fan is turned off, its angular speed decreases from \( 10 \) \( \text{rad/s} \) to \( 6.3 \) \( \text{rad/s} \) in \( 5.0 \) \( \text{s} \). What is the magnitude of the average angular acceleration of the fan?
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
\(0.4\,\text{rad/s}\)
\(0\,\text{m/s}^{2}\)
\(0\,\text{rad/s}^{2}\)
\(32\,\text{m/s}^{2}\)
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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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