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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. |
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In lacrosse, a typical throw is made by rotating the stick through an angle of roughly 90°, then releasing the ball when the stick is vertical, as shown above. If the 1 meter long stick is at rest when horizontal and the ball leaves the stick with a velocity of 10 m/s, what angular acceleration must the stick experience?
An object moves at a constant speed of [katex] 9.0 \frac{m}{s} [/katex] in a circular path of radius of 1.5 m. What is the angular acceleration of the object?
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during the entire 20-s interval? Assume constant angular acceleration while it is starting and stopping.
A rotating merry-go-round makes one complete revolution in 4.0 s. What is the linear speed and acceleration of a child seated 1.2 m from the center?
A system consists of two small disks, of masses m and 2m, attached to ends of a rod of negligible mass of length 3x. The rod is free to turn about a vertical axis through point P. The first mass, m, is located x away from point P, and therefore the other mass, of 2m, is 2x from point P. The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is . At time t = 0, the rod has an initial counterclockwise angular velocity ωi about P. The system is gradually brought to rest by friction.
Derive an expressions for the following quantities in terms of µ, m, x, g, and ωi.
(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 |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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