0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (a): Determine the final angular speed when power returns. | ||
| 1 | \( v_i = 500\,\text{rpm} \times \frac{2\pi\,\text{rad}}{60\,\text{s}} = \frac{500\times 2\pi}{60} = \frac{50\pi}{3}\,\text{rad/s} \) | Convert the initial speed from revolutions per minute to radians per second. |
| 2 | \( \Delta \theta = 200\,\text{rev} \times 2\pi = 400\pi\,\text{rad} \) | Calculate the total angular displacement during the 30 s power outage (each revolution is \(2\pi\) rad). |
| 3 | \( \Delta \theta = \frac{1}{2}(v_i + v_x)\,t \) | Use the kinematic equation for constant angular acceleration relating displacement, initial and final speeds over time \(t = 30\,s\). |
| 4 | \( v_x = \frac{2\Delta \theta}{t} – v_i = \frac{2(400\pi)}{30} – \frac{50\pi}{3} = \frac{800\pi}{30} – \frac{50\pi}{3} \) | Solve for the final angular speed \(v_x\) after 30 s. |
| 5 | \( \frac{800\pi}{30} = \frac{80\pi}{3}, \quad v_x = \frac{80\pi}{3} – \frac{50\pi}{3} = \frac{30\pi}{3} = 10\pi\,\text{rad/s} \) | Simplify the expression to obtain \(v_x\). This is equivalent to \(10\pi\,rad/s\) which can be converted to \(300\,rpm\) if desired. |
| 6 | \( \boxed{v_x = 10\pi\,\text{rad/s}} \) | Final answer for part (a): the flywheel spins at \(10\pi\,rad/s\) when power returns. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (b): Determine the time to stop and total revolutions if power did not return. | ||
| 1 | \( \alpha = \frac{v_x – v_i}{t} = \frac{10\pi – \frac{50\pi}{3}}{30} = \frac{\frac{30\pi – 50\pi}{3}}{30} = -\frac{20\pi}{90} = -\frac{2\pi}{9}\,\text{rad/s}^2 \) | Determine the constant angular deceleration \(\alpha\) using the change in angular velocity over 30 s. |
| 2 | \( 0 = v_i + \alpha T_{\text{stop}} \quad \Rightarrow \quad T_{\text{stop}} = -\frac{v_i}{\alpha} \) | Set the final angular velocity to zero to solve for the total stopping time \(T_{\text{stop}}\) from the moment of power failure. |
| 3 | \( T_{\text{stop}} = -\frac{\frac{50\pi}{3}}{-\frac{2\pi}{9}} = \frac{50\pi}{3} \times \frac{9}{2\pi} = 75\,\text{s} \) | Simplify to find that the flywheel takes 75 s to come to a complete stop. |
| 4 | \( \Delta \theta_{\text{total}} = \frac{v_i + 0}{2}T_{\text{stop}} = \frac{\frac{50\pi}{3}}{2} \times 75 = \frac{50\pi\times75}{6} = 625\pi\,\text{rad} \) | Calculate the total angular displacement using the average angular speed during deceleration. |
| 5 | \( \text{Revolutions} = \frac{625\pi}{2\pi} = 312.5\,\text{rev} \) | Convert radians to revolutions since \(2\pi\) rad correspond to one complete revolution. |
| 6 | \( \boxed{T_{\text{stop}} = 75\,\text{s} \quad \text{and} \quad \text{Total Revolutions} = 312.5\,\text{rev}} \) | Final answers for part (b): the flywheel stops in 75 s making a total of 312.5 revolutions. |
Just ask: "Help me solve this problem."
Which of the following must be zero if an object is spinning at a constant rate? There may be more than one right answer.
A pulley has an initial angular speed of \( 12.5 \) \( \text{rad/s} \) and a constant angular acceleration of \( 3.41 \) \( \text{rad/s}^2 \). Through what angle does the pulley turn in \( 5.26 \) \( \text{s} \)?
Consider a rigid body that is rotating. Which of the following is an accurate statement?
Four forces are exerted on a disk of radius \( R \) that is free to spin about its center, as shown above. The magnitudes are proportional to the length of the force vectors, where \( F_1 = F_4 \), \( F_2 = F_3 \), and \( F_1 = 2F_2 \). Which two forces combine to exert zero net torque on the disk?
When is the angular momentum of a system constant?
Part (a): \(\boxed{10\pi\,\text{rad/s}}\) (which is equivalent to 300 rpm).\nPart (b): \(\boxed{75\,\text{s}}\) to come to a complete stop with a total of \(\boxed{312.5\,\text{rev}}\).
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
