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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \omega = 100,000 \, \text{rpm} \) | Given the angular velocity of the flywheel. |
| 2 | \( \omega = 100,000 \times \frac{2\pi \, \text{rad} }{1 \, \text{rev}} \times \frac{1 \, \text{min} }{60 \, \text{s} } \) | Convert from revolutions per minute (rpm) to radians per second (rad/s). |
| 3 | \( \omega = \frac{100,000 \times 2\pi}{60} \, \text{rad/s} \) | Combine the conversion factors. |
| 4 | \( \omega \approx 10472 \, \text{rad/s} \) | Simplify the expression to get the angular velocity in rad/s. |
| 5 | \( r = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm} = 0.1 \, \text{m} \) | Calculate the radius of the flywheel and convert to meters. |
| 6 | \( v = \omega r \) | Use the formula for linear speed on the rim of a rotating object: \( v = \omega r \). |
| 7 | \( v = 10472 \, \text{rad/s} \times 0.1 \, \text{m} \) | Substitute the values for \( \omega \) and \( r \) into the formula. |
| 8 | \( v \approx 1047.2 \, \text{m/s} \) | Calculate the linear speed: the speed of a point on the rim of the flywheel is \( \boxed{1047.2 \, \text{m/s}} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \omega_i = 10472 \, \text{rad/s} \) | Initial angular velocity from part (a). |
| 2 | \( \omega_f = 0.6 \times 10472 \, \text{rad/s} \) | Angular velocity decreases by 40%, so the final angular velocity is 60% of the initial value. |
| 3 | \( \omega_f = 0.6 \times 10472 \, \text{rad/s} = 6283.2 \, \text{rad/s} \) | Calculate the final angular velocity. |
| 4 | \( \alpha = \frac{\Delta \omega}{\Delta t} \) | The formula for angular acceleration where \( \Delta \omega = \omega_f – \omega_i \) and \( \Delta t \) is the time interval. |
| 5 | \( \alpha = \frac{6283.2 \, \text{rad/s} – 10472 \, \text{rad/s}}{30 \, \text{s}} \) | Substitute the known values into the formula. |
| 6 | \( \alpha = \frac{-4188.8 \, \text{rad/s}}{30 \, \text{s}} \) | Simplify the numerator. |
| 7 | \( \alpha \approx -139.6 \, \text{rad/s}^2 \) | Calculate the angular acceleration, which is . The magnitude is \( \boxed{139.6 \, \text{rad/s}^2} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \) | Use the kinematic equation for angular displacement under constant angular acceleration. |
| 2 | \( \theta = 10472 \, \text{rad/s} \times 30 \, \text{s} + \frac{1}{2} \times (-139.6 \, \text{rad/s}^2) \times (30 \, \text{s})^2 \) | Substitute the known values into the formula. |
| 3 | \( \theta = 10472 \times 30 + \frac{1}{2} \times (-139.6) \times 900 \) | Simplify the expression. |
| 4 | \( \theta = 314160 – 62820 \) | Calculate the individual terms. |
| 5 | \( \theta = 251340 \, \text{rad} \) | Combine the results to get the total angular displacement in radians. |
| 6 | \( \text{Revolutions} = \frac{\theta}{2\pi} \) | Convert angular displacement from radians to revolutions. |
| 7 | \( \text{Revolutions} = \frac{251340}{2\pi} \) | Substitute the value of \( \theta \). |
| 8 | \( \text{Revolutions} \approx 40000 \) | Calculate the total number of revolutions. The rotor makes approximately \( \boxed{40000 \, \text{revolutions}} \) during these 30 seconds. |
Just ask: "Help me solve this problem."
Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at 30° to the plane of the door. Which force exerts the greater torque about the door hinge?
An object’s angular momentum changes by [katex] 10\,\text{kg-m}^2\text{/s} [/katex] in 2.0 s. What magnitude average torque acted on this object?
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 5-meter long ladder is leaning against a wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform and has a mass of 20 kg. A person of mass 80 kg is standing on the ladder at a distance of 4 meters from the bottom of the ladder. The ladder makes an angle of 60 degrees with the ground. What is the force exerted by the wall on the ladder?
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?
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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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