0 attempts
0% avg
UBQ Credits
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | Choose CCW as positive \[\;\] |
Fixing a sign convention allows all torques (axial components) to be added algebraically. |
| 2 | Geometry of the rigid pair Outer wheel: \(r_o = 3R\) Inner wheel: \(r_i = 2R\) |
The radii are read directly from the diagram (labels 3R and 2R on the spokes). |
| 3 | Force A (top, outer rim) \[\tau_A = r_o (2F)(-1)=-(3R)(2F)=-6FR\] |
The 2 F force is tangent to the rim and points rightward at the top. \(\mathbf r\) is upward, \(\mathbf F\) is rightward, so \(\mathbf r\times\mathbf F\) is into the page (CW, negative). |
| 4 | Force B (left, outer rim) \[\tau_B = r_o (F)(+1)=+(3R)(F)=+3FR\] |
The downward F on the left gives a CCW (positive) torque: \(\mathbf r\) is left, \(\mathbf F\) is down. |
| 5 | Force C (bottom, outer rim) \[\tau_C = r_o (F)(+1)=+(3R)(F)=+3FR\] |
At the bottom, \(\mathbf r\) points down, \(\mathbf F\) points right. The cross-product is out of the page (CCW, positive). |
| 6 | Net torque on outer wheel \[\tau_{\text{outer}}=\tau_A+\tau_B+\tau_C=-6FR+3FR+3FR=0\] |
The three torques on the 3R rim cancel exactly; the rigid pair therefore experiences no net moment from these three forces. |
| 7 | Force D (inner rim, bottom-right) \[\tau_D = r_i (F)(+1)=+(2R)(F)=+2FR\] |
The rightward F applied tangentially to the 2R rim gives a CCW torque of magnitude \(r_iF\). |
| 8 | Total torque on the two-wheel system \[\tau_{\text{net}} = \tau_{\text{outer}} + \tau_D = 0 + 2FR = 2FR\] |
Because the wheels are rigidly fastened, torques about the common axis simply add. |
| 9 | Magnitude \[|\tau_{\text{net}}| = 2FR\] |
The question asks for the magnitude; the sign (CCW) is therefore dropped. |
Just ask: "Help me solve this problem."
A uniform, rigid rod of length \( 2 \) \( \text{m} \) lies on a horizontal surface. One end of the rod can pivot about an axis that is perpendicular to the rod and along the plane of the page. A \( 10 \) \( \text{N} \) force is applied to the rod at its midpoint at an angle of \( 37^{\circ} \). A second force \( F \) is applied to the free end of the rod so that the rod remains at rest, as shown in the figure. The magnitude of the torque produced by force \( F \) is most nearly
A uniform rod of mass \( M_0 \) and length \( L \) is free to rotate about a pivot at its left end and is released from rest when the rod is \( 30^{\circ} \) below the horizontal, as shown in the figure. With respect to the pivot, the rod has rotational inertia \( I_0 = \dfrac{1}{3} M_0 L^2 \). Which of the following expressions correctly represents the magnitude of the net torque exerted on the rod about the pivot at the moment the rod is released?
What condition(s) are necessary for static equilibrium?
One end of a string is wrapped around a pulley that is free to rotate with negligible friction about an axle at its center. The other end of the string is attached to a block. The block is released from rest and moves downward with constant acceleration. Which of the following correctly indicates whether the amount of work done on the pulley by the string during each successive complete rotation remains constant or increases, and provides a valid justification?
In an experiment, an external torque is applied to the edge of a disk of radius \( 0.5 \) \( \text{m} \) such that the edge of the disk speeds up as it continues to rotate. The tangential speed as a function of time is shown for the edge of the disk. The rotational inertia of the disk is \( 0.125 \) \( \text{kg} \cdot \text{m}^2 \). Can a student use the graph and the known information to calculate the net torque exerted on the edge of the disk?
By continuing you (1) agree to our Terms of Use and Terms of Sale 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
We crafted THE Ultimate A.P Physics 1 course so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
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.
