0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0 – 40}{6 – 2} = \frac{-40}{4} = -10 \, \text{rad/s}^2\] | Calculate the angular acceleration between \( t = 2 \) and \( t = 6 \) seconds, where \( \omega \) changes from \( 40 \, \text{rad/s} \) to \( 0 \, \text{rad/s} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 2 | \[v = r \cdot \omega = 0.25 \times 60 = 15 \, \text{m/s}\] | Calculate the speed of a point on the rim at \( t = 1 \) second using angular velocity \( 60 \, \text{rad/s} \) and radius \( 0.25 \, \text{m} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 3 | \[\theta = \omega \cdot t = 60 \cdot 2 = 120 \, \text{rad}\] | Find the angular displacement in the first 2 seconds with constant angular velocity of \( 60 \, \text{rad/s} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 4 | \[\theta = \frac{1}{2} \times (60 + 0) \times 4 = 120 \, \text{rad}\] | Calculate the angular displacement from \( t = 2 \) to \( t = 6 \) seconds with initial and final velocities of \( 60 \) and \( 0 \), respectively. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 5 | \[\theta = \frac{1}{2} \times (0 + (-40)) \times 2 = -40 \, \text{rad}\] | Calculate the angular displacement from \( t = 6 \) to \( t = 8 \) seconds, where \( \omega \) changes linearly. |
| 6 | \[\theta = -40 + 0 \times 2 = -40 \, \text{rad}\] | Calculate the angular displacement from \( t = 8 \) to \( t = 10 \) seconds, with constant angular velocity of \( -40 \, \text{rad/s} \). |
| 7 | \[\text{Total } \theta = 120 + 120 – 40 – 80 = 120 \, \text{rad}\] | Sum all angular displacements from \( t = 0 \) to \( t = 10 \) seconds. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 8 | \[\Delta x = \theta \cdot r = 120 \cdot 0.25 = 30 \, \text{m}\] | Calculate the total displacement on the ground after 10 seconds with total angular displacement and wheel radius. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 9 | \[a_t = r \cdot \alpha = 0.25 \cdot (-10) = -2.5 \, \text{m/s}^2\] | Calculate the tangential acceleration at \( t = 4 \) seconds using the angular acceleration and radius. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 10 | \[a_t = r \cdot \alpha = 0.25 \cdot 0 = 0 \, \text{m/s}^2\] | The tangential acceleration is zero since angular acceleration is zero at \( t = 1 \) second. |
| 11 | \[a_c = \frac{v^2}{r} = \frac{(15)^2}{0.25} = 900 \, \text{m/s}^2\] | Calculate the centripetal acceleration at \( t = 1 \) second using the velocity and radius. |
Just ask: "Help me solve this problem."
A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?
A rod may freely rotate about an axis that is perpendicular to the rod and is along the plane of the page. The rod is divided into four sections of equal length of 0.2 m each, and four forces are exerted on the rod, as shown in the figure. Frictional forces are considered negligible. Which of the following describes an additional torque that must be applied in order to keep the rod from rotating?
A solid sphere ([katex] I = \frac{2}{5}MR^2[/katex]) and a solid cylinder ([katex] I = \frac{1}{2}MR^2[/katex]), both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is
Two masses, my = 32 kg and mg = 38 kg, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius R = 0.311 m and mass 3.1 kg. Initially my is on the ground and mg rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of me just before it strikes the ground. Assume the pulley bearing is frictionless.
When a fan is turned off, its angular speed decreases from 10 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the average angular acceleration of the fan?
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.
