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(A) \(\alpha\) between \(2\) and \(10\) seconds
| Derivation/Formula | Reasoning |
|---|---|
| \[\alpha = \frac{\Delta \omega}{\Delta t} = \frac{-30 – 60}{6 – 2} = \frac{-90}{4} = -22.5 \, \text{rad/s}^2\] | Calculate the angular acceleration between \( t = 2 \) and \( t = 6 \) seconds, where \( \omega \) changes from \( 60 \, \text{rad/s} \) to \( -30 \, \text{rad/s} \). |
(B) \(\v\) at \(1\) second
| Derivation/Formula | Reasoning |
|---|---|
| \[v = r \cdot \omega = 0.25 \times 60 = 15 \, \text{m/s}\] | Calculate the speed of a point on the rim at \( t = 1 \) second using angular velocity \( 60 \, \text{rad/s} \) and radius \( 0.25 \, \text{m} \). |
(C) \(\Delta \theta \) from \(0\) and \(2\) seconds
| Derivation/Formula | Reasoning |
|---|---|
| \[\theta = \omega \cdot t = 60 \cdot 2 = 120 \, \text{rad}\] | Find the angular displacement in the first 2 seconds with constant angular velocity of \( 60 \, \text{rad/s} \). This is the area under the line and bounded by the x-axis from \(0\) to \(2\) seconds. |
(D) \(\Delta \theta \) from \(2\) and \(4\) seconds
| Derivation/Formula | Reasoning |
|---|---|
| \[\theta = \frac{1}{2} \times (60 + 10) \times 2 = 70 \, \text{rad}\] | We can use an angular kinematic eqaution: \(\theta = \frac{1}{2} \times (\omega_f + \omega_i) \times t\)
Calculate the angular displacement from \( t = 2 \) to \( t = 4 \) seconds with initial and final velocities of \( 60 \) and \( 10 \), respectively. Alternatively you can use the area under the line as done in the previous problem. |
(E) \(\Delta \theta \) from \(0\) and \(10\) seconds (ANGULAR displacement)
| Derivation/Formula | Reasoning |
|---|---|
| \[\theta_{0-2} \, \theta_{2-6} \, \theta_{6-8} \, \theta_{8-10}\] | Break into multiple segments, and find the displacement for each using kinematic formulas OR area under the line for those segments. |
| \[\theta_{0-2} = 2 \times 60 = 120 \, \text{rad}\] | Calculate the angular displacement from \( t = 0 \) to \( t = 2 \) seconds using area. |
| \[\theta_{2-6} = \frac{1}{2}(\omega_f + \omega_i) \times \Delta t \]
\[ = \frac{1}{2}(-30+60) \times 4 = 60 \, \text{rad}\] |
Calculate the angular displacement from \( t = 2 \) to \( t = 6 \) seconds using kinematics. |
| \[\theta_{6-8} = -30 \times 2 = -60 \, \text{rad}\] | Calculate the angular displacement from \( t = 6 \) to \( t = 8 \) seconds using area. |
| \[\theta_{8-10} = \frac{1}{2} \times -30 \times 2 = -30 \, \text{rad}\] | Calculate the angular displacement from \( t = 8 \) to \( t = 10 \) seconds using area (of triangle). |
| \[\theta_{0-10} = 120 + 60 + 60 + (-60) + (-30) = 150 \, \text{rad}\] | Add up all the displacements from every segment. |
(F) \(\Delta \x \) from \(0\) and \(10\) seconds (LINEAR displacement)
| Derivation/Formula | Reasoning |
|---|---|
| \[\Delta x = 150 \times .25 = 37.5 \, \text{m}\] | Linear displacement is simply: \(\Delta x = \Delta \theta \times r \). We can use the angular displacement (\(\Delta \theta \)) we calculated from the previous problem. |
(G) \(\a_{tan}\) at \(4\) seconds
| Derivation/Formula | Reasoning |
|---|---|
| \[a_t = r \cdot \alpha = 0.25 \cdot (-22.5) = -5.6 \, \text{m/s}^2\] | Calculate the tangential acceleration at \( t = 4 \) seconds using the angular acceleration (found earlier in part a) and radius. |
(H) \(\a_{tan}\) and \(\a_{c}\) at \(1\) seconds
| Derivation/Formula | Reasoning |
|---|---|
| \[a_t = r \cdot \alpha = 0.25 \cdot 0 = 0 \, \text{m/s}^2\] | The tangential acceleration is zero since angular acceleration is zero at \( t = 1 \) second. |
| \[a_c = \frac{v^2}{r} = \frac{(15)^2}{0.25} = 900 \, \text{m/s}^2\] | Calculate the centripetal acceleration at \( t = 1 \) second using the linear velocity at \(1\) second. To find linear velocity use \(v = \omega \times r\) |
Just ask: "Help me solve this problem."
The rotating systems, shown in the figure above, differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left), or a distance r/2 from the axis of rotation (right). What happens if you release the hanging blocks simultaneously from rest?
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
In both cases, a massless rod is supported by fulcrum, and a 200-kg hanging mass is suspended from the left end of the rod by a cable. A downward force F keeps the rod in rest. The rod in Case A is 50 cm long, and the rod in case B is 40 cm long (each rod is marked at 10-cm intervals). The magnitude of each vertical force F exerted on the rod will be
A pulley has an initial angular speed of \( 12.5 \) \( \text{rad/s} \) and a constant angular acceleration of \( 3.41 \) \( \text{rad/s}^2 \). Through what angle does the pulley turn in \( 5.26 \) \( \text{s} \)?
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.
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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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