0 attempts
0% avg
UBQ Credits
(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."
Four systems are in rotational motion. Which of the following combinations of rotational inertia and angular speed for each of the systems corresponds to the greatest rotational kinetic energy?
| System | Rotational Inertia | Angular Speed |
|---|---|---|
| A | \( I_0 \) | \( \omega_0 \) |
| B | \( I_0 \) | \( 4\, \omega_0 \) |
| C | \( 2 I_0 \) | \( 2\, \omega_0 \) |
| D | \( 6 I_0 \) | \( \omega_0 \) |
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
A point on the edge of a disk rotates around the center of the disk with an initial angular velocity of 3 rad/s clockwise. The graph shows the point’s angular acceleration as a function of time. The positive direction is considered to be counterclockwise. All frictional forces are considered to be negligible.
A disk of radius 35 cm rotates at a constant angular velocity of 10 rad/s. How fast does a point on the rim of the disk travel (in m/s)?
During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which was used to create the graph that is shown. A frictional torque is exerted on the wheel. A student makes the following statement about the data. “The frictional torque exerted on the wheel is independent of the wheel’s angular speed.” Does the data from the graph support the student’s statement? Justify your selection.
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted the ultimate A.P Physics 1 course that simplifies everything so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
