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# Part (a): Solving Using Conservation of Energy
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] U_i + K_i = U_f + K_f [/katex] | The total mechanical energy (kinetic + potential) at the initial state equals the total mechanical energy at the final state due to conservation of energy. |
| 2 | [katex] U_i = m_g \cdot g \cdot h [/katex] | Initial potential energy is due to [katex] m_g [/katex] being at a height [katex] h [/katex] from the ground. [katex] m_y [/katex] is on the ground, so its potential energy is zero. |
| 3 | [katex] U_i = 38 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{m} [/katex] | Calculate the numerical value of [katex] U_i [/katex]. [katex] g [/katex] (acceleration due to gravity) is 9.8 m/s² and [katex] h [/katex] is 2.5 m. |
| 4 | [katex] U_i = 931 \, \text{J} [/katex] | Multiplying the values provides the initial potential energy. |
| 5 | [katex] K_i = 0 [/katex] | Initially, the system is at rest so the initial kinetic energy is zero. |
| 6 | [katex] K_f = \frac{1}{2}(m_y + m_g + m_p) v^2 [/katex] | Final kinetic energy calculated as the sum of the kinetic energies of [katex] m_g [/katex], [katex] m_y [/katex] and the pulley. [katex] m_p [/katex] is the mass of the pulley. The pulley contributes translational kinetic energy due to its mass. |
| 7 | [katex] U_f = 0 [/katex] | At the final state, [katex] m_g [/katex] is on the ground, so its potential energy is zero. |
| 8 | [katex] 931 \, \text{J} = \frac{1}{2} (32 \, \text{kg} + 38 \, \text{kg} + 3.1 \, \text{kg}) v^2 [/katex] | Use [katex] U_i + K_i = U_f + K_f [/katex] and solve for [katex] v [/katex]. |
| 9 | [katex] v = \sqrt{\frac{2 \times 931 \, \text{J}}{73.1 \, \text{kg}}} [/katex] | Simplify the equation to solve for [katex] v [/katex]. |
| 10 | [katex] v = \sqrt{25.48 \, \text{m}^2/\text{s}^2} [/katex] | Calculate the value under the square root. |
| 11 | [katex] v = 5.05 \, \text{m/s} [/katex] | The final velocity of [katex] m_g [/katex] just before it strikes the ground. |
# Part (b): Solving Using Forces, Torque, and Kinematics
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] F = m \cdot a [/katex] | Newton’s second law, the net force applied on the system causes acceleration. |
| 2 | [katex] F_{my} – F_{mg} = (m_y + m_g + m_p) \cdot a [/katex] | Forces due to weights of [katex] m_y [/katex] and [katex] m_g [/katex], including the pulley’s inertia, where [katex] F_{my} = m_y \cdot g [/katex] and [katex] F_{mg} = m_g \cdot g [/katex]. |
| 3 | [katex] T = \frac{m_g g – m_y g}{1 + \frac{m_p}{m_y + m_g}} [/katex] | Balance the forces taking into account the pulley rotation effect, where [katex] T [/katex] is the net torque and [katex] m_p [/katex] is moment of inertia derived as [katex] \frac{1}{2} m_p R^2 [/katex] assuming a uniform cylinder. |
| 4 | [katex] a = \frac{m_g g – m_y g}{m_y + m_g + \frac{1}{2} m_p} [/katex] | Solving the earlier equation for [katex] a [/katex], the acceleration of the system, where [katex] \frac{1}{2} m_p [/katex] comes from dividing the pulley’s moment of inertia by [katex] R^2 [/katex]. |
| 5 | [katex] v_f^2 = 2 \cdot a \cdot 2.5 \, \text{m} [/katex] | Use kinematic equation for final velocity, incorporating the calculated acceleration and distance fallen, which is 2.5 m. |
| 6 | [katex] v_f = \sqrt{2 \cdot a \cdot 2.5 \, \text{m}} [/katex] | Solve for final velocity using the value for [katex] a [/katex]. |
| 7 | Approximate numerical calculation similar to Part (a) | The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency. |
Just ask: "Help me solve this problem."
Two workers are holding a thin plate with length 5 m and height 2 m at rest by supporting the plate in the bottom corners. The workers are standing at rest on a slope of 10 degrees. Treat these supporting forces as vertical normal forces and calculate their magnitudes and state if both workers are sharing “the job” fairly.
A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above. The outside edge of the disk is moving at a linear speed \( v \), and the clay is moving at speed \( \frac{v}{2} \). The clay sticks to the outside edge of the disk. How does the angular momentum of the system after the clay sticks compare to the angular momentum of the system before the clay sticks, and what is an explanation for the comparison?
Angular momentum cannot be conserved if
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.
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to a rod of negligible mass of length \( 3l \) as shown above. The rod is free to turn about a vertical axis through 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 \( \omega_0 \) about \( P \). The system is gradually brought to rest by friction. Develop expressions for the following quantities in terms of \( \mu \), \( m \), \( l \), \( g \), and \( \omega_0 \).
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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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