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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\tau = FR[/katex] | The torque ([katex]\tau[/katex]) exerted on the cylinder is due to the force [katex]F[/katex] applied at a radius [katex]R[/katex]. The formula for torque is the force times the perpendicular distance (radius in this case) from the axis of rotation. |
| 2 | [katex]\tau = I\alpha[/katex] | Newton’s second law for rotation states that the torque is equal to the moment of inertia ([katex]I[/katex]) times the angular acceleration ([katex]\alpha[/katex]). |
| 3 | [katex]I = \frac{1}{2}MR^2[/katex] | The moment of inertia for a solid cylinder about its axis is given by this formula, where [katex]M[/katex] is the mass and [katex]R[/katex] is the radius of the cylinder. |
| 4 | [katex]FR = \frac{1}{2}MR^2 \alpha[/katex] | Substitute the moment of inertia of the cylinder into the torque equation. |
| 5 | [katex]\alpha = \frac{2F}{MR}[/katex] | Solve for the angular acceleration ([katex]\alpha[/katex]) by isolating [katex]\alpha[/katex] on one side of the equation. |
| 6 | [katex]\omega^2 = \omega_0^2 + 2\alpha \theta[/katex] | Use the kinematic equation for rotational motion to relate the angular displacement ([katex]\theta[/katex]) to the final angular velocity ([katex]\omega[/katex]). Here, [katex]\omega_0[/katex] (initial angular velocity) is zero as the cylinder starts from rest. |
| 7 | [katex]\omega^2 = 2\alpha \theta[/katex] | Substitute [katex]\omega_0 = 0[/katex] into the equation because the cylinder starts from rest. |
| 8 | [katex]\omega = \sqrt{2\alpha \theta} = \sqrt{\frac{4F\theta}{MR}}[/katex] | Substitute the value of [katex]\alpha[/katex] from Step 5 into the equation to find [katex]\omega[/katex]. |
| 9 | [katex]K = \frac{1}{2}I\omega^2[/katex] | The total kinetic energy ([katex]K[/katex]) of the rotating cylinder is given by the formula for rotational kinetic energy, where [katex]I[/katex] is the moment of inertia and [katex]\omega[/katex] is the angular velocity. |
| 10 | [katex]K = \frac{1}{2} \times \frac{1}{2}MR^2 \times \left(\frac{4F\theta}{MR}\right)[/katex] | Substitute the expressions for [katex]I[/katex] and [katex]\omega[/katex] into the kinetic energy formula. |
| 11 | [katex]K = \frac{F\theta R}{2}[/katex] | Simplify the equation to get the final expression for the kinetic energy. |
| 12 | [katex]K = \frac{F\theta R}{2}[/katex] | Conclude with the neat, simplified expression for the kinetic energy of the cylinder after it has rotated through an angle [katex]\theta[/katex]. |
Just ask: "Help me solve this problem."
A uniform stick has length \( L \). The moment of inertia about the center of the stick is \( I_0 \). A particle of mass \( M \) is attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
A car is moving up the side of a circular roller coaster loop of radius \( 12 \) \( \text{m} \). The angular velocity is \( 1.8 \) \( \text{rad/s} \) and angular acceleration is \( -0.82 \) \( \text{rad/s}^2 \). The car is at the same elevation as the center of the loop. Find the magnitude and direction (relative to the horizontal) of the acceleration.
Angular momentum cannot be conserved if
A 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
A 5-meter long ladder is leaning against a wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform and has a mass of 20 kg. A person of mass 80 kg is standing on the ladder at a distance of 4 meters from the bottom of the ladder. The ladder makes an angle of 60 degrees with the ground. What is the force exerted by the wall on the ladder?
[katex]K = \frac{F\theta R}{2}[/katex]
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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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