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Finding tension in the string
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]\tan(\theta) = \frac{R}{h}[/katex] | The angle [katex]\theta[/katex] is between the string and the vertical axis. [katex]h[/katex] is the vertical distance from the ball to the pivot. |
| 2 | [katex]h = L \cos(\theta)[/katex] | From the geometry of the cone. |
| 3 | [katex]\tan(\theta) = \frac{R}{L \cos(\theta)}[/katex] | Substituting the expression for [katex]h[/katex]. |
| 4 | [katex]\sin(\theta) = \frac{R}{\sqrt{R^2 + h^2}}[/katex] | From the right triangle formed by the string, [katex]h[/katex], and [katex]R[/katex]. |
| 5 | [katex]F_{T} \sin(\theta) = F_{c}[/katex] | The horizontal component of tension provides the centripetal force ([katex]F_{c}[/katex]). |
| 6 | [katex]F_{c} = m \frac{v^2}{R}[/katex] | Centripetal force formula. [katex]v[/katex] is the velocity of the ball. |
| 7 | [katex]F_{T} \cos(\theta) = mg[/katex] | The vertical component of tension balances gravity. |
| 8 | [katex]F_{T} = \frac{mg}{\cos(\theta)}[/katex] | Isolating [katex]F_{T}[/katex] in the vertical balance. |
| 9 | [katex]F_{T} = \frac{mg}{\cos(\theta)} = \frac{mg}{\sqrt{1 – \sin^2(\theta)}}[/katex] | Using [katex]\cos(\theta) = \sqrt{1 – \sin^2(\theta)}[/katex]. |
| 10 | [katex]F_{T} = \frac{mg}{\sqrt{1 – \left(\frac{R}{\sqrt{R^2 + h^2}}\right)^2}}[/katex] | Substituting [katex]\sin(\theta)[/katex]. |
| 11 | [katex]\boxed{F_{T} = \frac{mg}{\sqrt{1 – \frac{R^2}{R^2 + L^2 \cos^2(\theta)}}}}[/katex] | Final expression for tension, substituting [katex]h = L \cos(\theta)[/katex]. |
Finding period of the pendulum
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F_{c} = m \frac{v^2}{R}[/katex] | Centripetal force formula. |
| 2 | [katex]F_{T} \sin(\theta) = m \frac{v^2}{R}[/katex] | The horizontal component of tension provides the centripetal force. |
| 3 | [katex]v = R \omega[/katex] | Relationship between linear velocity and angular velocity ([katex]\omega[/katex]). |
| 4 | [katex]m \frac{(R \omega)^2}{R} = F_{T} \sin(\theta)[/katex] | Substituting [katex]v[/katex] with [katex]R \omega[/katex]. |
| 5 | [katex]\omega^2 = \frac{F_{T} \sin(\theta)}{mR}[/katex] | Isolating [katex]\omega^2[/katex]. |
| 6 | [katex]\omega = \sqrt{\frac{g}{R \tan(\theta)}}[/katex] | Using [katex]F_{T} \sin(\theta) = mg \sin(\theta)[/katex] and simplifying. |
| 7 | [katex]T = \frac{2\pi}{\omega}[/katex] | Period ([katex]T |
Just ask: "Help me solve this problem."
A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius R and is vertical. In terms of R and constants, find the minimum height h above the bottom of the loop the block must start from so it makes it around the loop.
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/min when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?
Two identical object rests on a platform rotating at constant speed. Object A is at distance of half the platform’s radius from the center. Object B lays at edge of the platform. Assuming the platform continues rotating at the same speed, how does the centripetal force of the two objects compare?
A car is safely negotiating an unbanked circular turn at a speed of 17 m/s on dry road. However, a long wet patch in the road appears and decreases the maximum static frictional force to one-fifth of its dry-road value. If the car is to continue safely around the curve, by what factor would the it need to change the original velocity?
A linear spring of negligible mass requires a force of \( 18.0 \, \text{N} \) to cause its length to increase by \( 1.0 \, \text{cm} \). A sphere of mass \( 75.0 \, \text{g} \) is then attached to one end of the spring. The distance between the center of the sphere \( M \) and the other end \( P \) of the un-stretched spring is \( 25.0 \, \text{cm} \). Then the sphere begins rotating at constant speed in a horizontal circle around the center \( P \). The distance \( P \) and \( M \) increases to \( 26.5 \, \text{cm} \).
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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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