0 attempts
0% avg
UBQ Credits
# Part (a): At what time does the motor reverse direction?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\omega = 20 – \frac{1}{2}t^2[/katex] | Given the angular velocity of the electric motor as a function of time. |
| 2 | [katex]\omega = 0[/katex] | To find the time when the motor reverses direction, set the angular velocity [katex]\omega[/katex] equal to zero. |
| 3 | [katex]0 = 20 – \frac{1}{2}t^2[/katex] | Substitute [katex]\omega = 0[/katex] into the given angular velocity equation. |
| 4 | [katex]\frac{1}{2}t^2 = 20 [/katex] | Rearrange the equation to solve for [katex]t^2[/katex]. |
| 5 | [katex]t^2 = 40[/katex] | Multiply both sides by 2 to isolate [katex]t^2[/katex]. |
| 6 | [katex]t = \sqrt{40}[/katex] | Solve for [katex]t[/katex] by taking the square root of both sides. Include only the positive root because time cannot be negative. |
| 7 | [katex]t = \boxed{6.32 \, \text{s}}[/katex] | Final answer: The motor reverses direction at [katex] t = 6.32 \, \text{s} [/katex]. |
# Part (b): Through what angle does the motor turn between [katex]t=0 \, \text{s}[/katex] and [katex]t=8 \, \text{s}[/katex]?
Note – If you haven’t learned calculus yet, plot the given \( \omega\) v \(t\) equation. Then approximate the area bound by the line and the x axis, up to the 8 second mark. This area represents the angular displacement of the motor.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\theta = \int_0^t \omega \, dt[/katex] | The angle [katex]\theta[/katex] through which the motor turns is the integral of angular velocity over time. |
| 2 | [katex]\theta = \int_0^8 \left(20 – \frac{1}{2} t^2 \right) dt [/katex] | Substitute the given angular velocity expression into the integral and set the limits from [katex] t = 0 \, \text{s} [/katex] to [katex] t = 8 \, \text{s} [/katex]. |
| 3 | [katex]\theta = \left[ 20t – \frac{1}{2} \cdot \frac{t^3}{3} \right]_0^8[/katex] | Integrate the expression term by term: [katex]\int 20 \, dt = 20t[/katex] and [katex]\int -\frac{1}{2}t^2 \, dt = -\frac{1}{2} \cdot \frac{t^3}{3}[/katex]. |
| 4 | [katex]\theta = \left[ 20t – \frac{t^3}{6} \right]_0^8 [/katex] | Simplify the integrated expression. |
| 5 | [katex]\theta = \left( 20 \times 8 – \frac{8^3}{6}\right) – \left( 20 \times 0 – \frac{0^3}{6} \right)[/katex] | Evaluate the expression at the limits: [katex] t = 8 \, \text{s} [/katex] and [katex] t = 0 \, \text{s} [/katex]. |
| 6 | [katex]\theta = \left( 160 – \frac{512}{6} \right) – 0[/katex] | Simplify and calculate the values. |
| 7 | [katex]\theta = 160 – 85.33[/katex] | Subtract the second term from the first term to find the angle. |
| 8 | [katex]\theta = \boxed{74.67 \, \text{rad}}[/katex] | The motor turns through an angle of [katex] 74.67 \, \text{rad} [/katex] between [katex] t = 0 \, \text{s} [/katex] and [katex] t = 8 \, \text{s} [/katex]. |
Just ask: "Help me solve this problem."
Pulleys \( X \) and \( Y \) are each attached to a block by a string that wraps around the pulley. Both blocks are released and have the same linear acceleration \( a \). As the blocks fall, the pulleys rotate about their centers. Pulley \( Y \) has a larger radius than Pulley \( X \). How does the angular acceleration \( \alpha_X \) of Pulley \( X \) compare to the angular acceleration \( \alpha_Y \) of Pulley \( Y \)?
A disk increases from 2 complete revolutions in 2 seconds to 5 complete revolutions in 2 seconds. What is its average angular acceleration?
A boy is sitting at a distance [katex] d_1 [/katex] from the fulcrum, and girl is sitting at a distance [katex] d_2 [/katex] from the fulcrum, with [katex] d_1 > d_2 [/katex]. The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
A meter stick with a uniformly distributed mass of \( 0.5 \) \( \text{kg} \) is supported by a pivot placed at the \( 0.25 \) \( \text{m} \) mark from the left. At the left end, a small object of mass \( 1.0 \) \( \text{kg} \) is placed at the zero mark, and a second small object of mass \( 0.5 \) \( \text{kg} \) is placed at the \( 0.5 \) \( \text{m} \) mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released.
A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
By continuing you (1) agree to our Terms of Sale and Terms of Use 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
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.
