| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 (a) | \( v = r \omega \) | For rolling motion, the linear velocity \( v \) at the bottom is related to the angular velocity \( \omega \) by the radius \( r \). Here, the radius \( r \) is 0.16 m. |
| 2 (a) | \( \omega = \frac{v}{r} \) | Calculate the angular velocity \( \omega \) at the bottom using the given linear velocity \( v = 3.2 \, \text{m/s} \). |
| 3 (a) | \( \omega = \frac{3.2 \, \text{m/s}}{0.16 \, \text{m}} = 20 \, \text{rad/s} \) | Substitute the values into the equation to find \( \omega \). |
| 4 (a) | \( \alpha = \frac{\omega}{t} \) | Angular acceleration \( \alpha \) is calculated using the angular velocity \( \omega \) and the time \( t \) it takes to reach that angular velocity. |
| 5 (a) | \( v^2 = v_0^2 + 2aL \) | Using the kinematic equation for linear motion, where \( a \) is linear acceleration and \( L = 1.5 \, \text{m} \) is the length of the incline. |
| 6 (a) | \( 3.2^2 = 0 + 2a \times 1.5 \) | \( a = \frac{(3.2)^2}{2 \times 1.5} = \frac{10.24}{3} \approx 3.413 \, \text{m/s}^2 \) |
| 7 (a) | \( a = r\alpha \) | Relate linear acceleration \( a \) to angular acceleration \( \alpha \). |
| 8 (a) | \( \alpha = \frac{a}{r} = \frac{3.413 \, \text{m/s}^2}{0.16 \, \text{m}} \approx 21.33 \, \text{rad/s}^2 \) | Compute angular acceleration \( \alpha \). |
| 1 (b) | \( \omega = \alpha t \) | Use the equation for angular velocity \( \omega \) related to angular acceleration \( \alpha \) and time \( t \). |
| 2 (b) | \( \omega_{\text{rpm}} = 7329 = \omega \frac{60}{2\pi} \) | Convert \( \omega \) from rpm to rad/s for calculation. |
| 3 (b) | \( \omega = 7329 \cdot \frac{2\pi}{60} \approx 767.23 \, \text{rad/s} \) | Find \( \omega \) in rad/s. |
| 4 (b) | \( t = \frac{\omega}{\alpha} = \frac{767.23}{419} \approx 1.83 \, \text{s} \) | Compute the time \( t \). |
| 1 (c) | \( \Delta \omega = \alpha \Delta t \) | Angular velocity change \( \Delta \omega \) is given, calculate the time \( \Delta t \) using angular acceleration \( \alpha \). |
| 2 (c) | \( \Delta \omega = 33.3 \, \text{rad/s} – 3.33 \, \text{rad/s} = 29.97 \, \text{rad/s} \) | Calculate the change in angular velocity \( \Delta \omega \). |
| 3 (c) | \( \Delta t = \frac{\Delta \omega}{\alpha} = \frac{29.97}{5.15} \approx 5.82 \, \text{s} \) | Calculate the time \( \Delta t \) needed to reach the final angular velocity. |
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At time \( t = 0 \), a disk starts from rest and begins spinning about its center with a constant angular acceleration of magnitude \( \alpha \). At time \( t_f \), the disk has angular speed \( \omega_f \). Which of the following expressions correctly compares the final angular displacement \( \theta_f \) of the disk at time \( t_f \) to the angular displacement \( \theta_{1/2} \) at time \( \frac{t_f}{2} \)?
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?
Old-fashioned clocks and watches have an hour hand, a minute hand and a second hand. What is the angular frequency of the second hand?
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 \( 0.300 \) \( \text{m} \)?
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? \( I_{sphere} = \frac{2}{5}MR^2\), \( I_{cylinder} = \frac{1}{2}MR^2\), \( I_{pipe} = MR^2\)
A grinding wheel is in the form of a uniform solid disk of radius \( 7.00 \) \( \text{cm} \) and mass \( 2.00 \) \( \text{kg} \). It starts from rest and accelerates uniformly under the action of the constant torque of \( 0.600 \) \( \text{N m} \) that the motor exerts on the wheel.
A 0.72-m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 N·m which accelerates it uniformly from rest through a total of 160 revolutions in 15.0 s. What is the mass of the sphere?
In an experiment, an external torque is applied to the edge of a disk of radius \( 0.5 \) \( \text{m} \) such that the edge of the disk speeds up as it continues to rotate. The tangential speed as a function of time is shown for the edge of the disk. The rotational inertia of the disk is \( 0.125 \) \( \text{kg} \cdot \text{m}^2 \). Can a student use the graph and the known information to calculate the net torque exerted on the edge of the disk?
A uniform solid cylinder of mass \( M \) and radius \( R \) 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 \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
(a) Angular acceleration of the ball is \( \approx 21.33 \, \text{rad/s}^2 \).
(b) Time for the CD player to reach full speed is \( \approx 1.83 \, \text{s} \).
(c) Time to accelerate for the rotating object is \( \approx 5.82 \, \text{s} \).
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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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