0 attempts
0% avg
UBQ Credits
Part A – Angular Speed
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T\cos(\theta)=mg\] | The vertical component of the tension \(T\) balances the gravitational force \(mg\) on the ball. |
| 2 | \[T\sin(\theta)=m\omega^2r\] | The horizontal component of \(T\) provides the centripetal force (\(m\omega^2r\)) needed for circular motion. |
| 3 | \[\frac{T\sin(\theta)}{T\cos(\theta)}=\frac{m\omega^2r}{mg}\] | Dividing the horizontal equation by the vertical one eliminates \(T\) to relate \(\omega\) and \(\theta\). |
| 4 | \[\tan(\theta)=\frac{\omega^2r}{g}\] | This simplifies the relationship between the angle \(\theta\) and the angular speed \(\omega\). |
| 5 | \[\omega^2=\frac{g\tan(\theta)}{r}\] | Solving for \(\omega^2\) in terms of \(\tan(\theta)\), \(r\), and \(g\). |
| 6 | \[\sin(\theta)=\frac{r}{\ell}\quad \text{and}\quad \cos(\theta)=\sqrt{1-\frac{r^2}{\ell^2}}=\frac{\sqrt{\ell^2-r^2}}{\ell}\] | Using the geometry of the conical pendulum, where the horizontal radius \(r\) relates to the string length \(\ell\) and angle \(\theta\). |
| 7 | \[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{r}{\sqrt{\ell^2-r^2}}\] | Expressing \(\tan(\theta)\) in terms of the given variables \(r\) and \(\ell\). |
| 8 | \[\omega^2=\frac{g}{r}\cdot\frac{r}{\sqrt{\ell^2-r^2}}=\frac{g}{\sqrt{\ell^2-r^2}}\] | Substituting the expression for \(\tan(\theta)\) into the equation for \(\omega^2\) simplifies the result. |
| 9 | \[\boxed{\omega=\sqrt{\frac{g}{\sqrt{\ell^2-r^2}}}}\] | Taking the square root yields the final expression for the angular speed \(\omega\) in terms of \(\ell\), \(r\), and \(g\). |
Part B – Tension
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T\cos(\theta)=mg\] | The vertical component of the tension balances the weight of the ball. |
| 2 | \[T=\frac{mg}{\cos(\theta)}\] | Solving for the tension \(T\) from the vertical equilibrium equation. |
| 3 | \[\cos(\theta)=\frac{\sqrt{\ell^2-r^2}}{\ell}\] | Expressing \(\cos(\theta)\) in terms of the string length \(\ell\) and the horizontal radius \(r\) using geometry. |
| 4 | \[T=\frac{mg}{\frac{\sqrt{\ell^2-r^2}}{\ell}}=\frac{mg\ell}{\sqrt{\ell^2-r^2}}\] | Substituting \(\cos(\theta)\) into the equation for \(T\) and simplifying. |
| 5 | \[\boxed{T=\frac{mg\ell}{\sqrt{\ell^2-r^2}}}\] | This is the final expression for the tension \(T\) in the string in terms of \(L\) (\(\ell\)), \(m\), \(r\), and \(g\). |
Just ask: "Help me solve this problem."
The coefficient of static friction between hard rubber and normal street pavement is about 0.85. On how steep a hill (maximum angle) can you leave a car parked?
A pendulum consists of a ball of mass [katex] m [/katex] suspended at the end of a massless cord of length [katex] L [/katex]. The pendulum is drawn aside through an angle of 60° with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
A 5.5 kg block slides down a 30º incline that is 2.2 m long. If µ = 0.20, what is the acceleration of the block?
A \( 4700 \, \text{kg} \) truck carrying a \( 900 \, \text{kg} \) crate is traveling at \( 25 \, \text{m/s} \) to the right along a straight, level highway, as shown above. The truck driver then applies the brakes, and as it slows down, the truck travels \( 55 \, \text{m} \) in the next \( 3.0 \, \text{s} \). The crate does not slide on the back of the truck.
A car travels to right at constant velocity. The net force on the car is
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.
