| 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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The object shown in the diagram below consists of a cylinder of mass \( 100 \) \( \text{kg} \) and radius \( 25.0 \) \( \text{cm} \) connected by four thin rods, each of mass \( 5.00 \) \( \text{kg} \) and length \( 0.75 \) \( \text{m} \), to a thin-outer ring of mass \( 20.0 \) \( \text{kg} \). A small chunk of metal of mass \( 1.00 \) \( \text{kg} \) is welded to the outer ring. Determine the moment of inertia of the entire assembly about the center of the inner cylinder, treating the metal chunk as a point mass. Hint: The moment of inertia of a disk about it center is \(\tfrac{1}{2} M R^2\), a thin rod about it center is \(\tfrac{1}{12}ML^2\), and a thin hoop about its center is \(I = MR^2\).
A high-speed flywheel in a motor is spinning at \( 500 \) \( \text{rpm} \) when a power failure suddenly occurs. The flywheel has a mass of \( 40 \) \( \text{kg} \) and a diameter of \( 75 \) \( \text{cm} \). The power is off for \( 30 \) \( \text{s} \) and during this time the flywheel slows due to friction in its axle bearings. During this time the flywheel makes \( 200 \) complete revolutions.
Consider a rigid body that is rotating. Which of the following is an accurate statement?
Old-fashioned clocks and watches have an hour hand, a minute hand and a second hand. What is the angular frequency of the second hand?
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 person’s center of mass is easily found by having the person lie on a reaction board. A horizontal, \( 2.3 \) \( \text{m} \)-long, \( 6.1 \) \( \text{kg} \) reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A \( 64 \) \( \text{kg} \) woman lies on the reaction board with her feet over the pivot. The scale reads \( 27 \) \( \text{kg} \). What is the distance from the woman’s feet to her center of mass? Express your answer with the appropriate units.
A 6.0-cm-diameter gear rotates with angular velocity \( \omega = \left(20-\frac {1}{2} t^2 \right) \, \text {rad/s} \), where \(t\) is in seconds. At \(t = 4.0 \, \text{s}\), what are
The elliptical orbit of a comet is shown above. Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2. At position 2, the comet’s kinetic energy is
At time \( t = 0 \), a disk starts from rest and begins spinning about its center with a constant angular acceleration of magnitude \( \alpha \). At time \( t_f \), the disk has angular speed \( \omega_f \). Which of the following expressions correctly compares the final angular displacement \( \theta_f \) of the disk at time \( t_f \) to the angular displacement \( \theta_{1/2} \) at time \( \frac{t_f}{2} \)?
\(\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;\text{(clockwise)},\;\tau_{\text{net},B}=0\;\text{Nm},\;\tau_{\text{net},C}=0\;\text{Nm}\)
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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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