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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 \(g\). |
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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