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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."
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
A \(2 \, \text{kg}\) ball is swung in a vertical circle. The length of the string the ball is attached to is \(0.7 \, \text{m}\). It takes \(0.4 \, \text{s}\) for the ball to travel one revolution (assume the ball travels at constant speed).
What is a man’s apparent weight at the equator if his weight is \(500 \, \text{N}\)? The Earth’s radius is \(6.37 \times 10^{6} \, \text{m}\).
A ball is attached to the end of a string. It is swung in a vertical circle of radius \( 2.5 \) \( \text{m} \). What is the minimum velocity that the ball must have to make it around the circle?
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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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