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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\omega_1 = 8\,\text{rev/s}\times 2\pi\,\text{rad/rev} = 16\pi\,\text{rad/s}\] | Convert the first angular speed to radians per second using \(2\pi\,\text{rad}=1\,\text{rev}\). |
| 2 | \[\omega_2 = 6\,\text{rev/s}\times 2\pi\,\text{rad/rev} = 12\pi\,\text{rad/s}\] | Convert the second angular speed to radians per second in the same way. |
| 3 | \[v_1 = r_1\,\omega_1 = 0.60\,\text{m}\times 16\pi = 9.6\pi\,\text{m/s}\] | Use \(v = r\omega\) to find the linear speed for the \(0.60\,\text{m}\) sling. |
| 4 | \[v_2 = r_2\,\omega_2 = 0.90\,\text{m}\times 12\pi = 10.8\pi\,\text{m/s}\] | Apply the same formula for the \(0.90\,\text{m}\) sling. |
| 5 | \[\boxed{v_2 > v_1}\] | Since \(10.8\pi>9.6\pi\), the \(6\,\text{rev/s}\) case gives the greater linear speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a_{c1}=r_1\,\omega_1^2\] | Centripetal acceleration is \(a_c = r\omega^2\). |
| 2 | \[a_{c1}=0.60\,(16\pi)^2 = 0.60\times256\pi^2 = 153.6\pi^2\,\text{m/s}^2\] | Substitute \(r_1=0.60\,\text{m}\) and \(\omega_1=16\pi\,\text{rad/s}\). |
| 3 | \[\boxed{a_{c1}\approx1.5\times10^{3}\,\text{m/s}^2}\] | Evaluate \(\pi^2\approx9.87\) to obtain a numerical value. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a_{c2}=r_2\,\omega_2^2\] | Use the same centripetal acceleration formula. |
| 2 | \[a_{c2}=0.90\,(12\pi)^2 = 0.90\times144\pi^2 = 129.6\pi^2\,\text{m/s}^2\] | Insert \(r_2=0.90\,\text{m}\) and \(\omega_2=12\pi\,\text{rad/s}\). |
| 3 | \[\boxed{a_{c2}\approx1.3\times10^{3}\,\text{m/s}^2}\] | Calculate numerically to compare with part (b). |
Just ask: "Help me solve this problem."
A uniform rod of length \( L \) and mass \( M \) is free to rotate about one end, as shown in the diagram. The free end is released from rest at a horizontal position, as shown. The pivot point is supported by a stand so that only the free end can move. The moment of inertia of a rod about its end is \(\tfrac{1}{3} M L^{2}\).
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A uniform copper disk of radius \( R \) has a moment of inertia \( I \) around an axis passing through the center of the disk perpendicular to its plane. If the radius of the disk were only \( \dfrac{R}{2} \), but the thickness were the same, what would be the moment of inertia in terms of \( I \)? Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
A centrifuge rotor rotating at \( 9200 \) \( \text{rpm} \) is shut off and is eventually brought uniformly to rest by a frictional torque of \( 1.20 \) \( \text{N} \cdot \text{m} \). If the mass of the rotor is \( 3.10 \) \( \text{kg} \) and it can be approximated as a solid cylinder of radius \( 0.0710 \) \( \text{m} \), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \( \frac{1}{2} m r^2 \).
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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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