| Step | Formula / Derivation | Reasoning |
|---|---|---|
| 1 | \( a_c = \frac{v^2}{R} \) | Centripetal acceleration formula, where \(v\) is the linear velocity and \(R\) is the radius of the circular path. |
| 2 | \( v = \frac{2\pi R}{T} \) | The linear velocity \(v\) is the circumference of the circle \((2\pi R)\) divided by the period \(T\). |
| 3 | \( a_c = \frac{\left(\frac{2\pi R}{T}\right)^2}{R} \) | Substitute the expression for \(v\) from step 2 into the formula from step 1. |
| 4 | \( a_c = \frac{4\pi^2 R^2}{T^2 R} \) | Square the velocity and distribute the square over the numerator. |
| 5 | \( a_c = \frac{4\pi^2 R}{T^2} \) | Cancel one \(R\) from the numerator and denominator. |
| 6 | \( \boxed{a_c = \frac{4\pi^2 R}{T^2}} \) | The magnitude of the centripetal acceleration in terms of \(R\) and \(T\). \(g\) is not required as it does not factor into the centripetal acceleration calculation for horizontal circular motion. The magnitude of the acceleration is therefore solely dependent on the radius of the circular path and the period of revolution. |
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Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star?
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 car travels at a constant speed around a circular track whose radius is \(2.6 \, \text{km}\). The car goes once around the track in \(360 \, \text{s}\). What is the magnitude of the centripetal acceleration of the car?
A \(5.0 \, \text{g}\) coin is placed \(15 \, \text{cm}\) from the center of a turntable. The coin has coefficients of static and kinetic friction of \(\mu_s = 0.80\) and \(\mu_k = 0.50\). The turntable slowly speeds up to \(60 \, \text{rpm}\). Does the coin slide off the turntable?
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?
The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of \(0.80 \, \text{m}\) above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is \(0.35 \, \text{m}\). The mass of the marble is \(0.050 \, \text{kg}\). Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.
A car is moving up the side of a circular roller coaster loop of radius \( 12 \) \( \text{m} \). The angular velocity is \( 1.8 \) \( \text{rad/s} \) and angular acceleration is \( -0.82 \) \( \text{rad/s}^2 \). The car is at the same elevation as the center of the loop. Find the magnitude and direction (relative to the horizontal) of the acceleration.
Why do pilots sometimes black out while pulling out at the bottom of a dive?
A particle of mass \(m\) slides down a fixed, frictionless sphere of radius \(R\), starting from rest at the top.
In terms of \(m\), \(g\), \(R\), and \(\theta\), determine each of the following for the particle while it is sliding on the sphere.
Two wires are tied to the \(500 \, \text{g}\) sphere as shown above. The sphere revolves in a horizontal circle at a constant speed of \(7.2 \, \text{m/s}\). What is the tension in the upper wire? What is the tension in the lower wire?
\( \boxed{a_c = \frac{4\pi^2 R}{T^2}} \)
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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] |
Metric Prefixes
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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