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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T\cos\theta = mg\] | Vertical acceleration is zero, so the vertical component of the tension balances the weight. |
| 2 | \[T = \frac{mg}{\cos\theta}\] | Solve the equilibrium equation for the tension \(T\). |
| 3 | \[\sin\theta = \frac{R}{L},\qquad \cos\theta = \sqrt{1-\frac{R^{2}}{L^{2}}}\] | Geometry of the conical pendulum: the string makes an angle \(\theta\) with the vertical, so \(R = L\sin\theta\). |
| 4 | \[\boxed{\displaystyle T = \frac{mg}{\sqrt{1-\frac{R^{2}}{L^{2}}}}}\] | Substitute the expression for \(\cos\theta\) from Step 3 into Step 2. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T\sin\theta = m\frac{v^{2}}{R}\] | The horizontal component of the tension provides the required centripetal force. |
| 2 | \[v^{2} = \frac{T R\sin\theta}{m}\] | Solve Step 1 for \(v^{2}\). |
| 3 | \[v^{2} = \frac{(mg/\cos\theta)\,R\sin\theta}{m}=gR\tan\theta\] | Substitute \(T=mg/\cos\theta\); the mass \(m\) cancels. |
| 4 | \[v = \sqrt{gR\tan\theta}\] | Take the square root to find the speed. |
| 5 | \[P = \frac{2\pi R}{v}\] | The period is the circumference divided by the speed. |
| 6 | \[P = \frac{2\pi R}{\sqrt{gR\tan\theta}} = 2\pi \sqrt{\frac{R}{g\tan\theta}}\] | Insert \(v\) from Step 4 and simplify correctly (retain one factor of \(R\) 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 \(\tan\theta\) using the geometry from Part (a). |
| 8 | \[P = 2\pi \sqrt{\frac{R}{g\left(\frac{R}{\sqrt{L^{2}-R^{2}}}\right)}} = 2\pi \sqrt{\frac{\sqrt{L^{2}-R^{2}}}{g}}\] | Substitute \(\tan\theta\) from Step 7 into Step 6 and cancel \(R\). |
| 9 | \[P = 2\pi \sqrt{\frac{L\cos\theta}{g}}\] | Since \(\sqrt{L^{2}-R^{2}} = L\cos\theta\), this is an equivalent compact form. |
| 10 | \[\boxed{\displaystyle P = 2\pi \sqrt{\frac{L\sqrt{1-\frac{R^{2}}{L^{2}}}}{g}}}\] | Express the period solely in terms of \(L\), \(R\), and \(g\). |
Just ask: "Help me solve this problem."
A communications satellite orbits the Earth at an altitude of \(35{,}000 \, \text{km}\) above the Earth’s surface. Take the mass of Earth to be \(6 \times 10^{24} \, \text{kg}\) and the radius of Earth to be \(6.4 \times 10^6 \, \text{m}\). What is the satellite’s velocity?
An Olympic bobsled team goes through a horizontal curve at a speed of \( 120 \) \( \text{km/hr} \). If the radius of curvature is \( 10.0 \) \( \text{m} \), what is the apparent weight the crew experiences—express in terms of \( mg \)?
A ball of mass \( m \) is fastened to a string. The ball swings at constant speed in a vertical circle of radius \( R \) with the other end of the string held fixed. Neglecting air resistance, what is the difference between the string’s tension at the bottom of the circle and at the top of the circle?
Why do you tend to slide across the car seat when the car makes a sharp turn?
An 80 kg person sits in a swing that goes around in a circle. The chain connecting the swing to the center of the ride is 8 m long and it makes and angle of 40° with the horizontal. What is the speed of the person going around in a circle?
\(T = \frac{mg}{\sqrt{1 – \frac{R^{2}}{L^{2}}}}\)
\(P = 2\pi \sqrt{\frac{L\sqrt{1 – \frac{R^{2}}{L^{2}}}}{g}}\)
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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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