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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."
A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at [katex] 470 \, \frac{rad}{s^2}[/katex]. The ball is released after [katex]\frac{3}{4} [/katex] of a revolution. How high does it go above the center of the wheel?
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
A boy is sitting at a distance [katex] d_1 [/katex] from the fulcrum, and girl is sitting at a distance [katex] d_2 [/katex] from the fulcrum, with [katex] d_1 > d_2 [/katex]. The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on
A uniform solid sphere of mass M and radius R is placed on a frictionless horizontal surface. A massless string is wrapped around the sphere and is pulled with a force F. The string makes an angle of θ with the horizontal. What is the minimum value of the coefficient of static friction between the sphere and the surface required for the sphere to start rolling without slipping?
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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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