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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \tau_1 = F_1 \times R \] | The torque from force \( F_1 \) is calculated by the product of the force and the radius, acting perpendicular to the radius. |
| 2 | \[ \tau_2 = F_2 \times \frac{R}{2} \] | Torque from force \( F_2 \) using its radius of action which is \( \frac{R}{2} \). |
| 3 | \[ \tau_3 = F_3 \times R \] | Torque from force \( F_3 \) using full radius \( R \) as it acts perpendicular. |
| 4 | \[ \tau_4 = F_4 \times R \times \sin(30^\circ) \] | The torque from \( F_4 \) is calculated as it has a component \( \sin(30^\circ) \) perpendicular to the radius. |
| 5 | \[ F_1 = 2F_2 \] | Given relationship, so \( \tau_1 = 2F_2 R \). |
| 6 | \[ \tau_1 = 4F_2 \times \frac{R}{2} = 4 \tau_2 \] | Substitute \( F_1 = 2F_2 \) and compare torques \( \tau_1 \) and \( \tau_2 \). |
| 7 | \[ \tau_3 = F_2 \times R \] | Use \( F_3 = F_2 \), so torques are equal with opposite signs based on direction. |
| 8 | \[ \tau_4 = 0 \] | Not needed as it adds to \( \tau_2 \) but doesn’t balance others. |
| 9 | \[ \boxed{F_1, \ F_3} \] | Forces \( F_1 \) and \( F_3 \) cancel each other because their torques are equal and opposite. |
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The figure above shows a uniform beam of length \( L \) and mass \( M \) that hangs horizontally and is attached to a vertical wall. A block of mass \( M \) is suspended from the far end of the beam by a cable. A support cable runs from the wall to the outer edge of the beam. Both cables are of negligible mass. The wall exerts a force \( F_w \) on the left end of the beam. For which of the following actions is the magnitude of the vertical component of \( F_w \) smallest?
Five forces act on a rod that is free to pivot at point \( P \), as shown in the figure. Which of these forces is producing a counter-clockwise torque about point \( P \)?
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
A light string is attached to a massive pulley of known rotational inertia \( I_P \), as shown in the figure. A student must determine the relationship between the torque exerted on the pulley and the change in the pulley’s angular velocity when the torque is applied for \( 2.0 \) \( \text{s} \). In addition to a stopwatch to measure the time interval, what two measurements could the student make in order to determine the relationship? Select two answers.
What is the net torque acting on the pivot supporting a \(10 \, \text{kilogram}\) beam \(2 \, \text{meters}\) long as shown above? Assume that the positive direction is clockwise.
A man with mass \( m \) is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius \( R \) that rotates without friction at a constant angular velocity \( \omega \). Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.
Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since \( L \propto I \), the angular momentum will decrease proportionately and the platform will slow down.
Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.
Explain what is correct and incorrect about each students statement if anything.
A uniform, solid, \( 100 \) \( \text{kg} \) cylinder with a diameter of \( 1.0 \) \( \text{m} \) is mounted so it is free to rotate about a fixed, horizontal, frictionless axis that passes through the centers of its circular ends. A \( 10 \) \( \text{kg} \) block is hung from a very light, thin cord wrapped around the cylinder’s circumference. When the block is released, the cord unwinds and the block accelerates downward. What is the acceleration of the block?
A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting?
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 \( 8.0 \) \( \text{s} \)?
A force of \(17 \, \text{N}\) is applied to the end of a \(0.63 \, \text{m}\) long torque wrench at an angle \(45^\circ\) from a line joining the pivot point to the handle. What is the magnitude of the torque about the pivot point produced by this force?
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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] |
Metric Prefixes
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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