| Derivation or Formula | Reasoning |
|---|---|
| \[\text{Given points: }A=(0,0),\;F_1\text{ at }(0,-2),\;F_2\text{ at }(2,-2),\;F_3\text{ at }(3,-1)\] | Set the pivot at \(A\) and list each force’s application point. |
| \[\text{Let each force magnitude be }F\] | All three forces have equal magnitude \(F\). |
| \[\text{Use }\tau_z = r_x F_y – r_y F_x\] | In 2D, the torque about the pivot (out of page \(z\)-component) is given by \(\tau_z\). Positive corresponds to counterclockwise (CCW). |
| \[\vec r_1=(0,-2)-(0,0)=(0,-2)\] | Position vector from \(A\) to where \(F_1\) acts. |
| \[\vec F_1=\left(-\frac{F}{\sqrt2},-\frac{F}{\sqrt2}\right)\] | “SW at \(45^\circ\) to vertical” means left and down with equal components: \(F_x<0,\;F_y<0\), each of magnitude \(F/\sqrt2\). |
| \[\tau_{1A}=r_{1x}F_{1y}-r_{1y}F_{1x}=(0)\left(-\frac{F}{\sqrt2}\right)-(-2)\left(-\frac{F}{\sqrt2}\right)=-\frac{2F}{\sqrt2}=-\sqrt2F\] | Compute torque from \(F_1\) about \(A\). Negative means clockwise (CW). |
| \[\vec r_2=(2,-2)-(0,0)=(2,-2)\] | Position vector from \(A\) to where \(F_2\) acts. |
| \[\vec F_2=\left(-\frac{F}{\sqrt2},\frac{F}{\sqrt2}\right)\] | “NW at \(45^\circ\) to vertical” means left and up with equal components: \(F_x0\). |
| \[\tau_{2A}=r_{2x}F_{2y}-r_{2y}F_{2x}=2\left(\frac{F}{\sqrt2}\right)-(-2)\left(-\frac{F}{\sqrt2}\right)=\frac{2F}{\sqrt2}-\frac{2F}{\sqrt2}=0\] | Torque from \(F_2\) cancels because the moment from its vertical component is canceled by the moment from its horizontal component about \(A\). |
| \[\vec r_3=(3,-1)-(0,0)=(3,-1)\] | Position vector from \(A\) to where \(F_3\) acts. |
| \[\vec F_3=(F,0)\] | \(F_3\) acts horizontally east: \(F_x=F,\;F_y=0\). |
| \[\tau_{3A}=r_{3x}F_{3y}-r_{3y}F_{3x}=3(0)-(-1)(F)=F\] | The line of action is above/below pivot by \(1\,\text{m}\) (since \(r_y=-1\)), producing a CCW (positive) torque. |
| \[\tau_{\text{net},A}=\tau_{1A}+\tau_{2A}+\tau_{3A}=-\sqrt2F+0+F=(1-\sqrt2)F\] | Sum the three torques about \(A\). |
| \[\boxed{\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;<0\;\Rightarrow\;\text{CW}}\] | Since \(\sqrt2\approx1.414\), \((1-\sqrt2)F<0\), so the net torque about \(A\) is clockwise. |
| Derivation or Formula | Reasoning |
|---|---|
| \[B=(0,-1)\] | Set the pivot at \(B\). |
| \[\tau_z = r_x F_y – r_y F_x\] | Same torque relation about the new pivot. |
| \[\vec r_1=(0,-2)-(0,-1)=(0,-1)\] | Position vector from \(B\) to where \(F_1\) acts. |
| \[\vec F_1=\left(-\frac{F}{\sqrt2},-\frac{F}{\sqrt2}\right)\] | Same force components as before. |
| \[\tau_{1B}=0\left(-\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=-\frac{F}{\sqrt2}\] | Torque from \(F_1\) about \(B\) is negative (CW). |
| \[\vec r_2=(2,-2)-(0,-1)=(2,-1)\] | Position vector from \(B\) to where \(F_2\) acts. |
| \[\vec F_2=\left(-\frac{F}{\sqrt2},\frac{F}{\sqrt2}\right)\] | Same \(F_2\) components as before. |
| \[\tau_{2B}=2\left(\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=\frac{2F}{\sqrt2}-\frac{F}{\sqrt2}=\frac{F}{\sqrt2}\] | Net torque from \(F_2\) about \(B\) is positive (CCW). |
| \[\vec r_3=(3,-1)-(0,-1)=(3,0)\] | Position vector from \(B\) to where \(F_3\) acts. |
| \[\vec F_3=(F,0)\] | Same horizontal force. |
| \[\tau_{3B}=3(0)-0(F)=0\] | Since \(\vec r_3\) is purely horizontal and \(\vec F_3\) is also horizontal, the line of action passes through the pivot height, giving zero moment arm about \(B\). |
| \[\tau_{\text{net},B}=\left(-\frac{F}{\sqrt2}\right)+\left(\frac{F}{\sqrt2}\right)+0=0\] | Sum of torques cancels exactly. |
| \[\boxed{\tau_{\text{net},B}=0\;\text{Nm}\;\Rightarrow\;\text{zero}}\] | Equal and opposite torques from \(F_1\) and \(F_2\) cancel, and \(F_3\) produces no torque about \(B\). |
| Derivation or Formula | Reasoning |
|---|---|
| \[C=(1,-1)\] | Set the pivot at \(C\). |
| \[\tau_z = r_x F_y – r_y F_x\] | Same torque formula. |
| \[\vec r_1=(0,-2)-(1,-1)=(-1,-1)\] | Position vector from \(C\) to where \(F_1\) acts. |
| \[\vec F_1=\left(-\frac{F}{\sqrt2},-\frac{F}{\sqrt2}\right)\] | Same \(F_1\) components. |
| \[\tau_{1C}=(-1)\left(-\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=\frac{F}{\sqrt2}-\frac{F}{\sqrt2}=0\] | About \(C\), the perpendicular effects of \(F_1\)’s components cancel, giving zero net torque from \(F_1\). |
| \[\vec r_2=(2,-2)-(1,-1)=(1,-1)\] | Position vector from \(C\) to where \(F_2\) acts. |
| \[\vec F_2=\left(-\frac{F}{\sqrt2},\frac{F}{\sqrt2}\right)\] | Same \(F_2\) components. |
| \[\tau_{2C}=1\left(\frac{F}{\sqrt2}\right)-(-1)\left(-\frac{F}{\sqrt2}\right)=\frac{F}{\sqrt2}-\frac{F}{\sqrt2}=0\] | Similarly, \(F_2\)’s component torques cancel about \(C\). |
| \[\vec r_3=(3,-1)-(1,-1)=(2,0)\] | Position vector from \(C\) to where \(F_3\) acts. |
| \[\vec F_3=(F,0)\] | Horizontal eastward force. |
| \[\tau_{3C}=2(0)-0(F)=0\] | \(\vec r_3\) is horizontal and \(\vec F_3\) is horizontal, so there is no moment arm about \(C\). |
| \[\tau_{\text{net},C}=0+0+0=0\] | All three individual torques are zero about \(C\). |
| \[\boxed{\tau_{\text{net},C}=0\;\text{Nm}\;\Rightarrow\;\text{zero}}\] | The lines of action produce no net rotational tendency about \(C\); each force has either zero moment arm or cancels via its components. |
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A motorcycle has tires with a diameter of \( 44.0 \) \( \text{cm} \). Cruising down the highway, they are rotating at \( 1150 \) \( \text{rpm} \) (revolutions per minute).
Angular momentum cannot be conserved if
A car accelerates from \( 0 \) to \( 25 \) \( \text{m/s} \) in \( 5 \) \( \text{s} \). If the car’s tires have a diameter of \( 70 \) \( \text{cm} \), how many revolutions does a tire make while accelerating?
An airliner arrives at the terminal, and the engines are shut off. The rotor of one of the engines has an initial clockwise angular velocity of \( 2000 \) \( \text{rad/s} \). The engine’s rotation slows with an angular acceleration of magnitude \( 80.0 \) \( \text{rad/s}^2 \).
Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
A rod is initially at rest on a rough horizontal surface. Three forces are exerted on the rod with the magnitudes and directions shown in the figure. The force exerted in the center of the rod is an equidistant 0.5 m from both ends of the rod. If friction between the rod and the table prevents the rod from rotating, what is the magnitude of the torque exerted on the rod about its center from frictional forces?
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at either end of the rod?
A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
Two uniform solid balls, one of radius \( R \) and mass \( M \), the other of radius \( 2R \) and mass \( 8M \), roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
Find the following three values using just rotational kinematics.
\(\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;\text{(clockwise)},\;\tau_{\text{net},B}=0\;\text{Nm},\;\tau_{\text{net},C}=0\;\text{Nm}\)
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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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