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| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ T\cos\theta = mg \] | This equation comes from vertical force equilibrium; the vertical component of the tension \(T\) must balance the gravitational force \(mg\). |
| 2 | \[ T = \frac{mg}{\cos\theta} \] | We solve the vertical equilibrium equation for \(T\) by dividing both sides by \(\cos\theta\). |
| 3 | \[ \sin\theta = \frac{R}{L} \quad \text{and} \quad \cos\theta = \sqrt{1-\frac{R^2}{L^2}} \] | Since the ball moves in a horizontal circle, the radius is given by \(R = L\sin\theta\). We then use the Pythagorean identity \(\cos\theta = \sqrt{1-\sin^2\theta}\) to express \(\cos\theta\) in terms of \(R\) and \(L\). |
| 4 | \[ \boxed{T = \frac{mg}{\sqrt{1-\frac{R^2}{L^2}}}} \] | Substitute the expression for \(\cos\theta\) into the formula for \(T\) to obtain the tension in terms of \(m\), \(g\), \(R\), and \(L\). |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ T\sin\theta = m\frac{v^2}{R} \] | This equation represents the horizontal force balance. The horizontal component of the tension \(T\) provides the centripetal force \(m\frac{v^2}{R}\) required for circular motion. |
| 2 | \[ v^2 = \frac{T R \sin\theta}{m} \] | We solve the horizontal force equation for \(v^2\) by isolating it on one side. |
| 3 | \[ v^2 = \frac{(mg/\cos\theta) R \sin\theta}{m} = gR\tan\theta \] | Substitute \(T = \frac{mg}{\cos\theta}\) from the vertical balance and simplify; notice that \(m\) cancels out. |
| 4 | \[ v = \sqrt{gR\tan\theta} \] | Take the square root to find the speed \(v\) of the ball. |
| 5 | \[ P = \frac{2\pi R}{v} \] | The period \(P\) is found by dividing the circumference of the circle \(2\pi R\) by the speed \(v\) of the ball. |
| 6 | \[ P = \frac{2\pi R}{\sqrt{gR\tan\theta}} = 2\pi \sqrt{\frac{R}{gR\tan\theta}} = 2\pi \sqrt{\frac{1}{g\tan\theta}} \] | Simplify the expression by canceling \(R\) under the square root. |
| 7 | \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{R/L}{\sqrt{1-\frac{R^2}{L^2}}} = \frac{R}{\sqrt{L^2-R^2}} \] | Express \(\tan\theta\) in terms of \(R\) and \(L\) using \(\sin\theta = \frac{R}{L}\) and \(\cos\theta = \sqrt{1-\frac{R^2}{L^2}}\). |
| 8 | \[ P = 2\pi \sqrt{\frac{1}{g\left(\frac{R}{\sqrt{L^2-R^2}}\right)}} = 2\pi \sqrt{\frac{\sqrt{L^2-R^2}}{gR}} \] | Substitute the expression for \(\tan\theta\) into the period formula. |
| 9 | \[ P = \frac{2\pi R}{\sqrt{gR\tan\theta}} = 2\pi\sqrt{\frac{L\cos\theta}{g}} \] | An alternative derivation: using \(R = L\sin\theta\), one can show that \(P = 2\pi\sqrt{\frac{L\cos\theta}{g}}\). Since \(\cos\theta = \sqrt{1-\frac{R^2}{L^2}}\), this gives the same result. |
| 10 | \[ \boxed{P = 2\pi \sqrt{\frac{L\sqrt{1-\frac{R^2}{L^2}}}{g}}} \] | This is the final expression for the period \(P\) expressed in terms of \(L\), \(R\), and \(g\). |
Just ask: "Help me solve this problem."
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
A 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.
An airplane can safely bank when subjected to a centripetal acceleration of 8 g’s. If the airplane flies at a constant speed of 400 m/s, how long does it take to make a 180° turn?
Two satellites of equal mass, \( S_1 \) and \( S_2 \), orbit the Earth. \( S_1 \) is orbiting at a distance \( r \) from the Earth’s center at speed \( v \). \( S_2 \) orbits at a distance \( 2r \) from the Earth’s centre at speed \( \dfrac{v}{\sqrt{2}} \). The ratio of the centripetal force on \( S_1 \) to the centripetal force on \( S_2 \) is
A simple pendulum consists of a bob of mass 1.8 kg attached to a string of length 2.3 m. The pendulum is held at an angle of 30° from the vertical by a light horizontal string attached to a wall, as shown above.
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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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