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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 pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) starts to move downwards. Assuming that the coefficient of kinetic friction between the pulley and the rope is \( 0.2 \), and neglecting air resistance, determine
A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope.
What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?
Note: [katex] I_\text{disk} = \frac{1}{2}mr^2 [/katex]
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end, and the other endpoint is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the magnitude and direction of the force between the cable and the wall and of the force between the steel beam and the wall.
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.
Two points, A and B, are on a disk that rotates about an axis. Point A is \( 3 \) times as far from the axis as point B. If the speed of point B is \( v \), then what is the speed of point A?
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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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