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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| General set-up | \(\begin{aligned} \text{Let}&:\;T_y&=T\sin\theta \,\;(\text{vertical component of tension})\\ H_y&=\text{vertical component of wall force we seek}\\ a&=\text{block position ( }a=L\text{ or }L/2)\\ b&=\text{cable position ( }b=L\text{ or }L/2)\;. \end{aligned}\) | All four arrangements differ only in the values of a and b. The beam’s own weight \(Mg\) acts at \(L/2\). Horizontal components are irrelevant for the vertical reaction. |
| Static equilibrium conditions | Torque about wall hinge: \(T_y\,b = Mg\,\frac{L}{2}+Mg\,a\) Vertical forces: \(H_y + T_y -2Mg = 0\Rightarrow H_y = 2Mg – T_y\) |
Taking torques about the hinge eliminates the unknown horizontal wall force. The horizontal component of tension passes through the hinge and produces no torque, so only \(T_y\) appears. |
| (A) Cable at \(L\), Block at \(L\) | \(\begin{aligned} T_y &= Mg\,\frac{\dfrac{L}{2}+L}{L}=1.5\,Mg\\ H_y &= 2Mg-1.5Mg = 0.5\,Mg \end{aligned}\) | The upward lift supplied by the cable is \(1.5Mg\); the wall therefore needs to supply only \(0.5Mg\) upward. |
| (B) Cable at \(\tfrac{L}{2}\), Block at \(L\) | \(\begin{aligned} T_y &= Mg\,\frac{\dfrac{L}{2}+L}{\tfrac{L}{2}} = 3\,Mg\\ H_y &= 2Mg-3Mg = -1\,Mg\quad\Rightarrow\;|H_y| = 1\,Mg \end{aligned}\) | The cable lifts more than the total weight, so the wall must push downward with magnitude \(Mg\). |
| (C) Cable at \(L\), Block at \(\tfrac{L}{2}\) | \(\begin{aligned} T_y &= Mg\,\frac{\dfrac{L}{2}+\tfrac{L}{2}}{L}=1\,Mg\\ H_y &= 2Mg-1Mg = 1\,Mg \end{aligned}\) | With the block moved inward the cable lifts less, so the wall must supply \(Mg\) upward. |
| (D) Cable at \(\tfrac{L}{2}\), Block at \(\tfrac{L}{2}\) | \(\begin{aligned} T_y &= Mg\,\frac{\dfrac{L}{2}+\tfrac{L}{2}}{\tfrac{L}{2}} = 2\,Mg\\ H_y &= 2Mg-2Mg = 0 \end{aligned}\) | The cable’s upward force exactly balances the two weights, leaving the wall with no vertical reaction. |
| Comparison | \(|H_y|_{(A)} = 0.5Mg,\; |H_y|_{(B)} = Mg,\; |H_y|_{(C)} = Mg,\; |H_y|_{(D)} = 0\) | The smallest magnitude of the wall’s vertical force is zero, obtained in option (D). |
Just ask: "Help me solve this problem."
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 \(350\ \text{g}\) ball is attached to the end of a thin, uniform rod of mass \(500\ \text{g}\) and length \(1.2\ \text{m}\). The system is rotated in a horizontal circle about the opposite end of the rod. Calculate the moment of inertia of the system about the axis of rotation. Hint: the moment of inertia of a thin rod about the end of the rod is \(I = \tfrac{1}{3} m L^2\).
Which of the following situations will increase the moment of inertia of a solid cylinder \( I = \tfrac{1}{2} M R^{2} \) by the same amount?
A ice skater that is spinning in circles has an initial rotational inertia Ii. You can approximate her shape to be a cylinder. She is spinning with velocity ωi. As she extends her arms she 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.
What condition(s) are necessary for static equilibrium?
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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] |
General Metric Conversion Chart
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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