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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\rho = R – r\] | Because the ball rolls on the inside of the track, its center is offset inward by its radius. Thus, the center follows a circular path of radius \(\rho = R – r\). |
| 2 | \[\Delta h = R – r\] | At the vertical edge the ball’s center is at a height of \(0\) (relative to the track center chosen so that the lowest point is \(- (R – r)\)) and at the lowest point it is at \(y = -(R – r)\); hence the drop in height is \(R – r\). |
| 3 | \[m g (R – r) = \frac{1}{2} m v_x^2 + \frac{1}{2} I \omega^2\] | Applying conservation of energy: the loss in gravitational potential energy equals the sum of translational and rotational kinetic energies. |
| 4 | \[I = \frac{2}{5} m r^2 \quad \text{and} \quad v_x = \omega r\] | For a solid spherical ball, the moment of inertia is \(\frac{2}{5} m r^2\) and rolling without slipping implies \(v_x = \omega r\). |
| 5 | \[\frac{1}{2} m v_x^2 + \frac{1}{2}\left(\frac{2}{5}m r^2\right)\left(\frac{v_x}{r}\right)^2 = \frac{1}{2} m v_x^2 + \frac{1}{5} m v_x^2 = \frac{7}{10}m v_x^2\] | Substitute the moment of inertia and the no-slip condition to express the rotational kinetic energy in terms of \(v_x\), then combine both kinetic energies. |
| 6 | \[m g (R – r) = \frac{7}{10} m v_x^2\] | Set the gravitational potential energy lost equal to the total kinetic energy gained. |
| 7 | \[v_x^2 = \frac{10}{7} g (R – r)\] | Simplify by canceling \(m\) and solving for \(v_x^2\). |
| 8 | \[\boxed{v_x = \sqrt{\frac{10}{7} g (R – r)}}\] | Take the square root to obtain the final expression for the ball’s speed at the lowest point. |
Just ask: "Help me solve this problem."
A 2 kg model rocket is launched with a thrust force of 275 N and reaches a height of 90 m, moving at 150 m/s at its peak. What is the average air resistance force acting on the rocket during its ascent?
What condition(s) are necessary for static equilibrium?
A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?
A person holds a book at rest a few feet above a table. The person then lowers the book at a slow constant speed and places it on the table. Which of the following accurately describes the change in the total mechanical energy of the Earth–book system?
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.
\(\boxed{v_x = \sqrt{\frac{10}{7} g (R – r)}}\)
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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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