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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."
Consider a rigid body that is rotating. Which of the following is an accurate statement?
An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about the axis of rotation?
A disk of radius \( R = 0.5 \) \( \text{cm} \) rests on a flat, horizontal surface such that frictional forces are considered to be negligible. Three forces of unknown magnitude are exerted on the edge of the disk, as shown in the figure. Which of the following lists the essential measuring devices that, when used together, are needed to determine the change in angular momentum of the disk after a known time of \( 5.0 \) \( \text{s} \)?
An isolated spherical star of radius [katex] R_o [/katex], rotates about an axis that passes through its center with an angular velocity of [katex] \omega_o [/katex]. Gravitational forces within the star cause the star’s radius to collapse and decrease to a value [katex] r_o <R_o [/katex], but the mass of the star remains constant. A graph of the star’s angular velocity as a function of time as it collapses is shown. Which of the following predictions is correct about the angular momentum [katex] L [/katex] of the star immediately after the collapse?
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
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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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