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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."
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.
Select all that applies. If a constant net torque is applied to an object, that object will
A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of \( 2\pi \, \text{rad/s}^2 \). If the wheel has a radius of \( 2 \, \text{m} \), what distance does the wheel travel in \( 3 \) seconds?
A Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
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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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