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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 (a) | [katex] v = r \omega [/katex] | For rolling motion, the linear velocity [katex] v [/katex] at the bottom is related to the angular velocity [katex] \omega [/katex] by the radius [katex] r [/katex]. Here, the radius [katex] r [/katex] is 0.16 m. |
| 2 (a) | [katex] \omega = \frac{v}{r} [/katex] | Calculate the angular velocity [katex] \omega [/katex] at the bottom using the given linear velocity [katex] v = 3.2 \, \text{m/s} [/katex]. |
| 3 (a) | [katex] \omega = \frac{3.2 \, \text{m/s}}{0.16 \, \text{m}} = 20 \, \text{rad/s} [/katex] | Substitute the values into the equation to find [katex] \omega [/katex]. |
| 4 (a) | [katex] \alpha = \frac{\omega}{t} [/katex] | Angular acceleration [katex] \alpha [/katex] is calculated using the angular velocity [katex] \omega [/katex] and the time [katex] t [/katex] it takes to reach that angular velocity. |
| 5 (a) | [katex] v^2 = v_0^2 + 2aL [/katex] | Using the kinematic equation for linear motion, where [katex] a [/katex] is linear acceleration and [katex] L = 1.5 \, \text{m} [/katex] is the length of the incline. |
| 6 (a) | [katex] 3.2^2 = 0 + 2a \times 1.5 [/katex] | [katex] a = \frac{(3.2)^2}{2 \times 1.5} = \frac{10.24}{3} \approx 3.413 \, \text{m/s}^2 [/katex] |
| 7 (a) | [katex] a = r\alpha [/katex] | Relate linear acceleration [katex] a [/katex] to angular acceleration [katex] \alpha [/katex]. |
| 8 (a) | [katex] \alpha = \frac{a}{r} = \frac{3.413 \, \text{m/s}^2}{0.16 \, \text{m}} \approx 21.33 \, \text{rad/s}^2 [/katex] | Compute angular acceleration [katex] \alpha [/katex]. |
| 1 (b) | [katex] \omega = \alpha t [/katex] | Use the equation for angular velocity [katex] \omega [/katex] related to angular acceleration [katex] \alpha [/katex] and time [katex] t [/katex]. |
| 2 (b) | [katex] \omega_{\text{rpm}} = 7329 = \omega \frac{60}{2\pi} [/katex] | Convert [katex] \omega [/katex] from rpm to rad/s for calculation. |
| 3 (b) | [katex] \omega = 7329 \cdot \frac{2\pi}{60} \approx 767.23 \, \text{rad/s} [/katex] | Find [katex] \omega [/katex] in rad/s. |
| 4 (b) | [katex] t = \frac{\omega}{\alpha} = \frac{767.23}{419} \approx 1.83 \, \text{s} [/katex] | Compute the time [katex] t [/katex]. |
| 1 (c) | [katex] \Delta \omega = \alpha \Delta t [/katex] | Angular velocity change [katex] \Delta \omega [/katex] is given, calculate the time [katex] \Delta t [/katex] using angular acceleration [katex] \alpha [/katex]. |
| 2 (c) | [katex] \Delta \omega = 33.3 \, \text{rad/s} – 3.33 \, \text{rad/s} = 29.97 \, \text{rad/s} [/katex] | Calculate the change in angular velocity [katex] \Delta \omega [/katex]. |
| 3 (c) | [katex] \Delta t = \frac{\Delta \omega}{\alpha} = \frac{29.97}{5.15} \approx 5.82 \, \text{s} [/katex] | Calculate the time [katex] \Delta t [/katex] needed to reach the final angular velocity. |
Just ask: "Help me solve this problem."
A disk of radius 35 cm rotates at a constant angular velocity of 10 rad/s. How fast does a point on the rim of the disk travel (in m/s)?
How long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
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?
Two masses, my = 32 kg and mg = 38 kg, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius R = 0.311 m and mass 3.1 kg. Initially my is on the ground and mg rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of me just before it strikes the ground. Assume the pulley bearing is frictionless.
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
(a) Angular acceleration of the ball is [katex] \approx 21.33 \, \text{rad/s}^2 [/katex].
(b) Time for the CD player to reach full speed is [katex] \approx 1.83 \, \text{s} [/katex].
(c) Time to accelerate for the rotating object is [katex] \approx 5.82 \, \text{s} [/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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