| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\tau = F r\] | Torque \(\tau\) about the pivot equals the applied force \(F\) times the perpendicular lever arm \(r\). All forces are perpendicular to the bar, so \(\tau\) is simply the product \(F r\). |
| 2 | \[\tau_a = F L\] | Case (a): a single upward force \(F\) is applied at the right end of the bar (distance \(L\) from the pivot). |
| 3 | \[\tau_b = 2F \left(\tfrac{3}{4}L\right)=\tfrac{3}{2}F L\] | Case (b): the upward force \(2F\) is applied roughly three-quarters along the bar (visual estimate \(r=\tfrac{3}{4}L\)). Multiplying gives \(1.5\,F L\). |
| 4 | \[\tau_c = \tfrac{F}{2}\left(\tfrac{3}{4}L\right)+\tfrac{F}{2}(L)=\tfrac{7}{8}F L\] | Case (c): two equal upward forces \(\tfrac{F}{2}\) act at \(\tfrac{3}{4}L\) and \(L\). Adding the two torques gives \(0.875\,F L\). |
| 5 | \[\tau_d = \tfrac{F}{2}(L) – F\left(\tfrac{3}{4}L\right)= -\tfrac{1}{4}F L\] | Case (d): the upward force \(\tfrac{F}{2}\) (counter-clockwise) at \(L\) competes with the downward force \(F\) (clockwise) at \(\tfrac{3}{4}L\), giving a small clockwise net torque of magnitude \(0.25\,F L\). |
| 6 | \[|\tau_b| > |\tau_a| > |\tau_c| > |\tau_d|\] | Ordering the magnitudes shows case (b) has the largest torque, so the bar’s angular acceleration \(\alpha = \tau/I\) (and hence its angular speed) rises fastest there. |
| 7 | \[\boxed{\text{Case (b)}}\] | Therefore the angular speed increases most rapidly in case (b). |
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A massless rigid rod of length \(3d\) is pivoted at a fixed point \(W\), and two forces each of magnitude \(F\) are applied vertically upward as shown above. A third vertical force of magnitude \(F\) may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude \(F\) is applied at point?
A car is moving up the side of a circular roller coaster loop of radius \( 12 \) \( \text{m} \). The angular velocity is \( 1.8 \) \( \text{rad/s} \) and angular acceleration is \( -0.82 \) \( \text{rad/s}^2 \). The car is at the same elevation as the center of the loop. Find the magnitude and direction (relative to the horizontal) of the acceleration.
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
When a fan is turned off, its angular speed decreases from \( 10 \) \( \text{rad/s} \) to \( 6.3 \) \( \text{rad/s} \) in \( 5.0 \) \( \text{s} \). What is the magnitude of the average angular acceleration of the fan?
What is the ratio of the moment of inertia of a cylinder of mass \( m \) and radius \( r \) to the moment of inertia of a hoop of the same mass and same radius?
Two points, A and B, are on a disk that rotates about an axis. Point A is \( 3 \) times as far from the axis as point B. If the speed of point B is \( v \), then what is the speed of point A?
To increase the moment of inertia of a body about an axis, you must
Wheels \( A \) and \( B \) are connected by a moving belt and are both free to rotate about their centers. The belt does not slip on the wheels. The radius of Wheel \( B \) is twice the radius of Wheel \( A \). Wheel \( A \) has constant angular speed \( \omega_A \) and Wheel \( B \) has constant angular speed \( \omega_B \). Which of the following correctly relates \( \omega_A \) and \( \omega_B \)?
Two identical solid disks, each of mass \( M \) and radius \( R \), are welded together so that they touch at exactly one point on their rims. Determine the moment of inertia of the combined object about an axis that is perpendicular to the plane of the disks and passes through their point of contact. Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
Two disks, A and B, each experience a net external torque that varies over an interval of \( 5 \) \( \text{s} \). Disk B has a rotational inertia that is twice that of Disk A. The graph shown represents the angular momentum of the two disks as functions of time between \( t = 0 \) \( \text{s} \) and \( t = 5 \) \( \text{s} \). The average magnitudes of the net torques exerted on disks A and B from \( t = 0 \) \( \text{s} \) to \( t = 5 \) \( \text{s} \) are \( \tau_A \) and \( \tau_B \), respectively. Which of the following expressions correctly relates the magnitudes of the average torques?
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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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