| 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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Consider a rigid body that is rotating. Which of the following is an accurate statement?
An ice skater that is spinning in circles has an initial rotational inertia \(I_i\). You can approximate her shape to be a cylinder. She is spinning with velocity \(\omega_i\). As she extends her arms, her rotational inertia changes by a factor of \(x\) and her angular velocity changes by a factor of \(y\). Which one of the following options best describe \(x\) and \(y\)?
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first? Hint the rotational inertia of a sphere about its center is \(I = \frac{2}{5}mR^{2}\) and the rotational inertia of a cylinder about its center is \(I = \frac{1}{2}mR^{2}\).
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?
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.
The downward motion of an elevator is controlled by a cable that unwinds from a cylinder of radius \( 0.20 \) \( \text{m} \). What is the angular velocity of the cylinder when the downward speed of the elevator is \( 1.2 \) \( \text{m/s} \)?
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 meter stick with a uniformly distributed mass of \(0.5 \, \text{kg}\) is supported by a pivot placed at the \(0.25 \, \text{m}\) mark from the left. At the left end, a small object of mass \(1.0 \, \text{kg}\) is placed at the zero mark, and a second small object of mass \(0.5 \, \text{kg}\) is placed at the \(0.5 \, \text{m}\) mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released.
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
Consider a solid uniform sphere of radius \(R\) and mass \(M\) rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?
\(\tau_{\text{net},A}=(1-\sqrt2)F\;\text{Nm}\;\text{(clockwise)},\;\tau_{\text{net},B}=0\;\text{Nm},\;\tau_{\text{net},C}=0\;\text{Nm}\)
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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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