0 attempts
0% avg
UBQ Credits
Before solving the question, we can find the radius of the ball using Pythagorean theorem to get .866 m. We can also use the trig to solve for the angle each rope makes with the horizontal (30° for both ropes).
Sum of Forces in the Horizontal Direction:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] \cos(30) = \frac{\sqrt{3}}{2} [/katex] | Cosine of [katex]30^\circ[/katex]. |
| 2 | [katex] F_{\text{centripetal}} = \frac{mv^2}{r} [/katex] | Centripetal force for circular motion. |
| 3 | [katex] T_1 \cos(\theta) + T_2 \cos(\theta) = \frac{mv^2}{r} [/katex] | Sum of horizontal components of tension equals centripetal force. |
| 4 | [katex] T_1 \frac{\sqrt{3}}{2} + T_2 \frac{\sqrt{3}}{2} = \frac{(0.5)(7.2)^2}{0.866} [/katex] | Substitute values for [katex]m[/katex], [katex]v[/katex], [katex]r[/katex], and [katex]\cos(\theta)[/katex]. |
| 5 | [katex] \frac{\sqrt{3}}{2}(T_1 + T_2) = 29.93 [/katex] | Calculate centripetal force and factor out [katex]\frac{\sqrt{3}}{2}[/katex]. |
Sum of Forces in the Vertical Direction:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] \sin(30) = \frac{1}{2} [/katex] | Sine of [katex]30^\circ[/katex]. |
| 2 | [katex] w = mg [/katex] | Weight of the sphere. |
| 3 | [katex] T_2 \sin(\theta) + mg – T_1 \sin(\theta) = 0 [/katex] | Vertical forces must balance: upward tensions and downward weight. |
| 4 | [katex] T_2 \frac{1}{2} + (0.5)(9.8) – T_1 \frac{1}{2} = 0 [/katex] | Substitute values for [katex]m[/katex], [katex]g[/katex], and [katex]\sin(\theta)[/katex]. |
| 5 | [katex] \frac{1}{2}(T_2 – T_1) + 4.9 = 0 [/katex] | Factor out [katex]\frac{1}{2}[/katex] and calculate weight. |
Solving for Tensions:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | Solve equations | Use the system of equations to solve for [katex]T_1[/katex] and [katex]T_2[/katex]. |
| 2 | [katex] T_1 \approx 22.18 \text{ N} [/katex] | Numerical solution for [katex]T_1[/katex]. |
| 3 | [katex] T_2 \approx 12.38 \text{ N} [/katex] | Numerical solution for [katex]T_2[/katex]. |
Final Tensions:
Just ask: "Help me solve this problem."
A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius \( R \) and is vertical. In terms of \( R \) and constants, find the minimum height \( h \) above the bottom of the loop the block must start from so it makes it around the loop.
An adult exerts a horizontal force on a swing that is suspended by a rope of length \( L \), holding it at an angle \( \theta \) with the vertical. The child in the swing has a weight \( W \) and dimensions that are negligible. In terms of \( W \) and \( \theta \), determine:
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter \(8.0 \, \text{m}\). The room is spinning horizontally about an axis through its center at a rate of \(45 \, \text{rev/min}\) when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?
The maximum acceleration a pilot can withstand without blacking out is about \( 7.0 \) \( g \). In an endurance test for a jet plane’s pilot, what is the maximum speed he can tolerate if he is spun in a horizontal circle of diameter \( 85 \) \( \text{m} \)?
Upper wire: \(22 \, \text{N}\)
Lower wire: \(12 \, \text{N}\)
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 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.
