| Derivation/Formula | Reasoning |
|---|---|
| \[N = \frac{1150}{60}\] | Convert the rotation rate from \(1150\,\text{rpm}\) to revolutions per second \(N\). |
| \[\omega = 2\pi N\] | Each revolution corresponds to \(2\pi\) radians, so angular velocity is \(2\pi N\). |
| \[\boxed{\omega = 1.20\times10^{2}\,\text{rad/s}}\] | Substituting \(N = 19.2\,\text{rev/s}\) gives \(\omega \approx 1.20\times10^{2}\,\text{rad/s}\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[f = N = \frac{1150}{60}\] | The frequency \(f\) equals the revolutions per second. |
| \[T = \frac{1}{f}\] | The period \(T\) is the reciprocal of the frequency. |
| \[\boxed{f = 1.92\times10^{1}\,\text{Hz}}\] | Numerical value of the frequency. |
| \[\boxed{T = 5.22\times10^{-2}\,\text{s}}\] | Numerical value of the period. |
| Derivation/Formula | Reasoning |
|---|---|
| \[r = \frac{D}{2} = \frac{0.44\,\text{m}}{2}\] | Radius is half of the \(0.44\,\text{m}\) diameter, so \(r = 0.22\,\text{m}\). |
| \[v = \omega r\] | Tangential speed \(v\) on a circle is angular speed times radius. |
| \[v = (1.20\times10^{2})(0.22)\] | Insert \(\omega\) and \(r\) to compute \(v\). |
| \[\boxed{v = 26.5\,\text{m/s}}\] | Calculated tangential speed in \(\text{m/s}\). |
| \[\boxed{v = 95.4\,\text{km/h}}\] | Convert by multiplying by \(3.6\) to obtain \(\text{km/h}\). |
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An object is experiencing a nonzero net force. Which of the following statements is most accurate?
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
A windmill blade with a rotational inertia of \( 6.0 \) \( \text{kg} \cdot \text{m}^2 \) has an initial angular velocity of \( 8 \) \( \text{rad/s} \) in the clockwise direction. It is then given an angular acceleration of \( 4 \) \( \text{rad/s}^2 \) in the clockwise direction for \( 10 \) seconds. What is the change in rotational kinetic energy of the blade over this time interval?
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of \( 5.0 \) \( \text{rev/s} \) in \( 8.0 \) \( \text{s} \). 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 \( 12.0 \) \( \text{s} \). 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 man with mass \( m \) is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius \( R \) that rotates without friction at a constant angular velocity \( \omega \). Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.
Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since \( L \propto I \), the angular momentum will decrease proportionately and the platform will slow down.
Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.
Explain what is correct and incorrect about each students statement if anything.
Young David experimented with slings before tackling Goliath. He found that he could develop an angular speed of \( 8.0 \) \( \text{rev/s} \) in a sling \( 0.60 \) \( \text{m} \) long. If he increased the length to \( 0.90 \) \( \text{m} \), he could revolve the sling only \( 6.0 \) times per second.
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
The system above is NOT balanced since \(m_2\) is twice the mass of \(m_1\). Which of the following changes would NOT balance the system so that there is 0 net torque? Assume the plank has no mass of its own.
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first? Hint the rotational inertia of a sphere about its center is \(I = \frac{2}{5}mR^{2}\) and the rotational inertia of a cylinder about its center is \(I = \frac{1}{2}mR^{2}\).
\(1.20 \times 10^{2}\,\text{rad/s}\)
\(1.92 \times 10^{1}\,\text{Hz}\)
\(5.22 \times 10^{-2}\,\text{s}\)
\(26.5\,\text{m/s}\)
\(95.4\,\text{km/h}\)
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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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