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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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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?
An old record player could bring a disk up to its \(45\) RPM speed in less than a second. If the same size disk can also be brought up to a speed of \(75\) RPM in about the same amount of time on another player. Compare the torques exerted by each record player.
The angular velocity of a rotating disk of radius \(20 \, \text{cm}\) increases from \(1 \, \text{rad/s}\) to \(3 \, \text{rad/s}\) in \(0.5 \, \text{s}\). What is the linear tangential acceleration of a point on the rim of the disk during this time interval?
Initially, a ball has an angular velocity of \( 5.0 \) \( \text{rad/s} \) counterclockwise. Some time later, after rotating through a total angle of \( 5.5 \) \( \text{radians} \), the ball has an angular velocity of \( 1.5 \) \( \text{rad/s} \) clockwise.
A uniform solid cylinder of mass [katex] M [/katex] and radius [katex] R [/katex] is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force [katex] F [/katex] , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle [katex] \theta [/katex], what is the kinetic energy of the cylinder in terms of [katex] F [/katex] and [katex] \theta [/katex]?
A race car travels in a circular track of radius \( 200 \) \( \text{m} \). If the car moves with a constant speed of \( 80 \) \( \text{m/s} \),
The driver of a car traveling at \( 30.0 \) \( \text{m/s} \) applies the brakes and undergoes a constant negative acceleration of \( 2.00 \) \( \text{m/s}^2 \). How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of \( 0.300 \) \( \text{m} \)?
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?
An airliner arrives at the terminal, and the engines are shut off. The rotor of one of the engines has an initial clockwise angular velocity of \( 2000 \) \( \text{rad/s} \). The engine’s rotation slows with an angular acceleration of magnitude \( 80.0 \) \( \text{rad/s}^2 \).
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.
(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] |
Metric Prefixes
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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