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To solve this problem, we need to find the position of the center of mass of the system since that is where the string should be attached for the bar to balance horizontally.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}\] | The formula for the center of mass \(x_{\text{cm}}\) is given by dividing the sum of moments (mass times its position) by the total mass. |
| 2 | \[x_{\text{cm}} = \frac{(2 \, \text{kg} \times 0 \, \text{m}) + (4 \, \text{kg} \times 0.75 \, \text{m}) + (4 \, \text{kg} \times 1.5 \, \text{m})}{2 \, \text{kg} + 4 \, \text{kg} + 4 \, \text{kg}}\] | Calculate the moment for each mass. Mass 1 is at position 0 m, Mass 2 is at the midpoint (0.75 m), and Mass 3 is at the far right (1.5 m). The total length of the bar is 1.5 m. |
| 3 | \[x_{\text{cm}} = \frac{(0) + (3 \, \text{kg} \cdot \text{m}) + (6 \, \text{kg} \cdot \text{m})}{10 \, \text{kg}}\] | Simplify the numerator by calculating the moments for the given positions and mass values. |
| 4 | \[x_{\text{cm}} = \frac{9 \, \text{kg} \cdot \text{m}}{10 \, \text{kg}}\] | Sum the moments in the numerator. |
| 5 | \[x_{\text{cm}} = 0.9 \, \text{m}\] | Divide by the total mass to find the center of mass’s position. |
The string should be attached at \(\boxed{0.9 \, \text{m}}\) from the far left side of the bar, which matches option (c).
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A pulley has an initial angular speed of \( 12.5 \) \( \text{rad/s} \) and a constant angular acceleration of \( 3.41 \) \( \text{rad/s}^2 \). Through what angle does the pulley turn in \( 5.26 \) \( \text{s} \)?
Two masses, \( m_1 \) and \( m_2 \), are suspended on either side of a pulley with a radius \( R \), as shown. The heavier mass, \( m_2 \), is initially held at rest above the ground by a distance \( h \) before being released. An student measures that it takes an amount of time \( t \) for the heavier mass to hit the ground after being released.
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?
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to ends of a rod of negligible mass of length \( 3x \). The rod is free to turn about a vertical axis through point \( P \). The first mass, \( m \), is located \( x \) away from point \( P \), and therefore the other mass, of \( 2m \), is \( 2x \) from point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_i \) about \( P \). The system is gradually brought to rest by friction.
Derive expressions for the following quantities in terms of \( \mu \), \( m \), \( x \), \( g \), and \( \omega_i \).
A wheel 31 cm in diameter accelerates uniformly from 240rpm to 360rpm in 6.8 s. How far will a point on the edge of the wheel have traveled in this time?
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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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