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| Step | Formula / Derivation | Reasoning |
|---|---|---|
| 1 | [katex] F_c = \frac{mv^2}{R} [/katex] | Centripetal force formula, where [katex] m [/katex] is mass, [katex] v [/katex] is speed, and [katex] R [/katex] is radius. |
| 2 | [katex] v = \frac{2\pi R}{T} [/katex] | Speed [katex] v [/katex] is the circumference [katex] 2\pi R [/katex] divided by period [katex] T [/katex]. |
| 3 | [katex] T = \frac{1}{f} [/katex] | Period [katex] T [/katex] is the reciprocal of frequency [katex] f [/katex]. |
| 4 | [katex] v = 2\pi R f [/katex] | Substituting [katex] T = \frac{1}{f} [/katex] into [katex] v = \frac{2\pi R}{T} [/katex]. |
| 5 | [katex] F_c [/katex] constant, [katex] R [/katex] decreases | If [katex] R [/katex] decreases and [katex] F_c [/katex] is constant, [katex] v^2 [/katex] must decrease to maintain the equation. |
| 6 | [katex] v [/katex] decreases | Speed [katex] v [/katex] must decrease because [katex] v^2 [/katex] has decreased. |
| 7 | [katex] f [/katex] increases | Since [katex] v [/katex] decreases but [katex] R [/katex] decreases more (being in the denominator), the overall effect is an increase in [katex] f [/katex]. |
The correct answer is:
c. [katex] v [/katex] decreases and [katex] f [/katex] increases
In step 2, the relationship is derived from the definition of linear speed in circular motion, where the distance traveled in one revolution is the circumference of the circle. Substituting the period [katex] T [/katex] with its reciprocal, the frequency [katex] f [/katex], directly gives us the relationship used in step 4. This process shows how the frequency of revolutions is inversely proportional to the period and directly proportional to the linear speed and radius of the path.
Phy can also check your working. Just snap a picture!
Which of the following do not affect the maximum speed that a car can drive in a circle? Choose both correct answers.
Two wires are tied to the 500 g sphere shown below. The sphere revolves in a horizontal circle at a constant speed of 7.2 m/s. What is the tension in the upper wire? What is the tension in the lower wire?
A car is moving up the side of a circular roller coaster loop of radius 12 m. The angular velocity is [katex] 1.8 \, \text{rad/s} [/katex] and angular acceleration is [katex] -0.82 \, \text{rad/s}^2 [/katex]. The car is at the same elevation as the center of the loop. Find the magnitude and direction of the acceleration.
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 communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be [katex]6 \times 10^{24} \text{ kg}[/katex] the the radius of Earth to be [katex]6.4 \times 10^6 \text{ m}[/katex]. What is the satellite’s velocity?
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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