0 attempts
0% avg
UBQ Credits
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \(I = I_0 + I_{\text{particle}}\) | The total moment of inertia of the combined system is the sum of the moment of inertia of the stick about its center and the moment of inertia of the attached particle. |
2 | \(I_{\text{particle}} = M \cdot d^2\) | The moment of inertia of a point mass about a rotation axis is given by the product of its mass and the square of the distance from the axis. |
3 | \(d = \frac{L}{2}\) | The distance from the center of the stick to the end where the particle is attached is half the length of the stick, as the rotation axis is at the center. |
4 | \(I_{\text{particle}} = M \cdot \left(\frac{L}{2}\right)^2\) | Substitute the expression for \(d\) into the formula for the particle’s moment of inertia. |
5 | \(I_{\text{particle}} = \frac{1}{4} ML^2\) | Simplify the expression by calculating \(\left(\frac{L}{2}\right)^2\). |
6 | \(I = I_0 + \frac{1}{4} ML^2\) | Add the moment of inertia of the particle to the moment of inertia of the stick to get the total moment of inertia. |
7 | (a) \(\boxed{I_0 + \frac{1}{4} ML^2}\) | This matches option (a), which is the correct answer among the given choices. |
Just ask: "Help me solve this problem."
What is the net torque acting on the pivot supporting a 10-kilogram beam 2 meters long as shown above? Assume that the positive direction is clockwise.
Three forces of equal magnitude are applied to a \( 3 \)-m by \( 2 \)-m rectangle. Force \( F_1 \) and \( F_2 \) act at \( 45^\circ \) angles to the vertical as shown, while \( F_3 \) acts horizontally.
A wheel of moment of inertia of \( 5.00 \) \( \text{kg} \cdot \text{m}^2 \) starts from rest and accelerates under a constant torque of \( 3.00 \) \( \text{N} \cdot \text{m} \) for \( 8.0 \) \( \text{s} \). What is the wheel’s rotational kinetic energy at the end of \( 8.0 \) \( \text{s} \)?
Select all that applies. If a constant net torque is applied to an object, that object will
Two masses, \( m_1 \) and \( m_2 \), are suspended on either side of a pulley with a radius \( R \), as shown. The heavier mass, \( m_2 \), is initially held at rest above the ground by a distance \( h \) before being released. An student measures that it takes an amount of time \( t \) for the heavier mass to hit the ground after being released.
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
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.
NEW! PHY instantly solves any question
🔥 Elite Members get up to 30% off Physics Tutoring
🧠 Learning Physics this summer? Try our free course.
🎯 Need exam style practice questions? We’ve got over 2000.