| Derivation or Formula | Reasoning |
|---|---|
| \[F = \frac{mv^2}{r}\] | This is the centripetal force formula for uniform circular motion where \(F\) is the centripetal force, \(m\) is the mass, \(v\) is the speed, and \(r\) is the radius of the circle. |
| \[v = \sqrt{\frac{Fr}{m}}\] | Rearranging the centripetal force formula shows that the speed \(v\) is proportional to the square root of the radius \(r\) when \(F\) and \(m\) are constant. Thus, reducing \(r\) causes \(v\) to decrease. |
| \[v = 2\pi r f\] | This relationship connects the speed \(v\) to the frequency \(f\) of revolution in circular motion. |
| \[f = \frac{v}{2\pi r}\] | Solving for the frequency \(f\) from \(v = 2\pi r f\) gives \(f\) in terms of \(v\) and \(r\). |
| \[f = \frac{1}{2\pi r}\sqrt{\frac{Fr}{m}} = \frac{\sqrt{F}}{2\pi\sqrt{m}}\frac{1}{\sqrt{r}}\] | Substituting the expression for \(v\) into \(f = \frac{v}{2\pi r}\) shows that \(f\) is inversely proportional to \(\sqrt{r}\). Therefore, if \(r\) decreases, \(f\) increases. |
| Final Answer: \(v\) decreases, \(f\) increases | The analysis reveals that reducing the radius while keeping the centripetal force constant results in a lower speed and a higher frequency. This makes option (d) correct. Options (a), (b), and (c) are incorrect because they either predict an increase in \(v\) or a decrease in \(f\), which contradicts the derived relationships. |
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An \(80 \, \text{kg}\) person sits on a swing ride that moves in a horizontal circle. The swing is suspended by a chain of length \(8 \, \text{m}\). While in motion, the chain makes a \(40^\circ\) angle with the horizontal. What is the speed of the person?
Which of the following do not affect the maximum speed that a car can drive in a circle? Choose all correct answers.
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 car rounds a curve at a steady \( 50 \) \( \text{km/h} \). If it rounds the same curve at a steady \( 70 \) \( \text{km/h} \), will its acceleration be any different?
A pendulum consists of a ball of mass \( m \) suspended at the end of a massless cord of length \( L \). The pendulum is drawn aside through an angle of \( 60^\circ \) with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
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?
Imagine a hypothetical planet that has two moons. Moon \(\#1\) is in a circular orbit of radius \(R\) and has a mass \(M\).
A speed skater goes around a turn that has a radius of \(31 \, \text{m}\). The skater has a speed of \(14 \, \text{m/s}\) and experiences a centripetal force of \(460 \, \text{N}\). What is the mass of the skater?
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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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