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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] F_{\text{net}} = T – mg [/katex] | Write the net force equation for the ball at the bottom of the circle. Here, [katex] T [/katex] is the tension in the string and [katex] mg [/katex] is the gravitational force acting downwards on the ball. |
| 2 | [katex] F_{\text{net}} = \frac{mv^2}{r} [/katex] | Use the centripetal force formula where [katex] m [/katex] is mass, [katex] v [/katex] is the speed, and [katex] r [/katex] is the radius of the circle. |
| 3 | [katex] \frac{mv^2}{r} = T – mg [/katex] | Set the net force equal to the centripetal force. This combines both equations from Steps 1 and 2. |
| 4 | [katex] T = \frac{mv^2}{r} + mg [/katex] | Rearrange the equation to solve for the tension [katex] T [/katex] in the string. |
| 5 | [katex] T = \frac{(2.0 \, \text{kg})(4.0 \, \text{m/s})^2}{0.65 \, \text{m}} + (2.0 \, \text{kg})(9.8 \, \text{m/s}^2) [/katex] | Substitute the known values: mass [katex] m = 2.0 \, \text{kg} [/katex], speed [katex] v = 4.0 \, \text{m/s} [/katex], radius [katex] r = 0.65 \, \text{m} [/katex], and gravitational acceleration [katex] g = 9.8 \, \text{m/s}^2 [/katex]. |
| 6 | [katex] T = \frac{(2.0)(16)}{0.65} + 19.6 [/katex] | Calculate intermediate values for clarity: [katex] 4.0^2 = 16 [/katex] and [katex] (2.0)(9.8) = 19.6 [/katex]. |
| 7 | [katex] T = \frac{32}{0.65} + 19.6 [/katex] | Continue simplifying the equation. |
| 8 | [katex] T = 49.23 + 19.6 [/katex] | Divide to find the centripetal force component: [katex] \frac{32}{0.65} = 49.23 [/katex]. |
| 9 | [katex] T = 68.83 \, \text{N} [/katex] | Sum the forces to arrive at the final tension in the string: [katex] 49.23 + 19.6 = 68.83 [/katex]. |
Just ask: "Help me solve this problem."
Two identical satellites are placed in orbit of two different planets. Satellite A orbits Mars, and Satellite B orbits Jupiter. The orbital speeds of each satellite are the same. Which satellite has a greater orbital radius?
A delivery truck is traveling north. It then turns along a leftward circular curve. This the packages in the truck to slide to the RIGHT. Which of the following is true of the net force on the packages as they are sliding?
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?
An object moves at constant speed in a circular path of radius \( r \) at a rate of \( 1 \) revolution per second. What is its acceleration in terms of \(r\)?
Why do pilots sometimes black out while pulling out at the bottom of a dive?
69 N
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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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