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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] v = R \omega [/katex] | The relationship between linear velocity [katex] v [/katex], radius [katex] R [/katex], and angular velocity [katex] \omega [/katex]. Here [katex] v [/katex] is the tangential speed at the equator. |
| 2 | [katex] \omega = \frac{2\pi}{T} [/katex] | Angular velocity [katex] \omega [/katex] is calculated as the ratio of [katex] 2\pi [/katex] (the angle for one full rotation in radians) to the period [katex] T [/katex] of one rotation. |
| 3 | [katex] \omega = \frac{2\pi \, \text{rad}}{1.0 \, \text{s}} = 2\pi \, \text{rad/s} [/katex] | Substitute the period [katex] T = 1.0 [/katex] s to find the angular velocity. |
| 4 | [katex] v = 10 \times 10^{3} \, \text{m} \cdot 2\pi \, \text{rad/s} [/katex] | Substitute the radius [katex] R = 10 \, \text{km} = 10 \times 10^{3} \, \text{m} [/katex] and the calculated angular velocity into the formula for linear speed. |
| 5 | [katex] v = 20\pi \times 10^{3} \, \text{m/s} [/katex] | Calculate the product to find the linear speed. |
| 6 | \[katex] v \approx 62,832 \, \text{m/s} [/katex]\ | Finding the numerical approximation for [katex] 20\pi \times 10^{3} [/katex] by substituting [katex] \pi \approx 3.14159 [/katex], giving the speed at the equator of the neutron star. |
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Find the escape speed from a planet of mass \(6.89 \times 10^{25} \, \text{kg}\) and radius \(6.2 \times 10^{6} \, \text{m}\).
A race car traveling at a constant speed of \( 50 \) \( \text{m/s} \) drives around a circular track that is \( 500 \) \( \text{m} \) in diameter. What is the magnitude of the acceleration of the car?
A 2.2 kg ball on the end of a 0.35 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 5.3 m/s. Find the tension in the string.
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 car of mass \( M \) moves around a circularly banked curve on a freeway off-ramp. The off-ramp has a radius of curvature \( R \) and is raised to an angle \( \theta \) from the horizontal. The road is slick, and friction is negligible.
A car with speed \( v \) and an identical car with speed \( 3v \) both travel the same circular section of an unbanked (flat) road. If the frictional force required to keep the faster car on the road without skidding is \( F \), then the frictional force required to keep the slower car on the road without skidding is
A loop-de-loop roller coaster has a radius of \( 30 \) \( \text{m} \). Determine the apparent weight a \( 500 \) \( \text{N} \) person will feel at the bottom of the loop while traveling at a speed of \( 25 \) \( \text{m/s} \) and at the top of the loop while traveling at a speed of \( 20 \) \( \text{m/s} \).
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
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?
Which of the following best explains why astronauts experience weightlessness while orbiting the earth?
62,832 m/s
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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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