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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 hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.
A wheel 31 cm in diameter accelerates uniformly from 240rpm to 360rpm in 6.8 s. How far will a point on the edge of the wheel have traveled in this time?
A boy is sitting at a distance [katex] d_1 [/katex] from the fulcrum, and girl is sitting at a distance [katex] d_2 [/katex] from the fulcrum, with [katex] d_1 > d_2 [/katex]. The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
A meter stick with a uniformly distributed mass of 0.5 kg is supported by a pivot placed at the 0.25 m mark from the left. At the left end, a small object of mass 1.0 kg is placed at the zero mark, and a second small object of mass 0.5 kg is placed at the 0.5 m mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released,.
A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?
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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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