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| Step | Formula and Calculations | Reasoning |
|---|---|---|
| 1 | [katex]F_{\text{static max}} = \mu_s \times m \times g[/katex] | Maximum static friction force (F_static max) is the product of the coefficient of static friction (μs), mass of the coin (m), and acceleration due to gravity (g). |
| 2 | [katex]m = 5.0 , \text{g} = 0.005 , \text{kg}[/katex] | Convert the mass of the coin from grams to kilograms. |
| 3 | [katex]F_{\text{static max}} = 0.80 \times 0.005 \times 9.81[/katex] | Substitute the values of μs, m, and g (9.81 m/s²) into the static friction formula. |
| 4 | [katex]\omega = \frac{60 \times 2 \pi}{60}[/katex] | Convert the angular velocity from rpm to radians per second (rad/s). |
| 5 | [katex]F_{\text{centripetal}} = m \times r \times \omega^2[/katex] | Centripetal force (F_centripetal) is calculated using the mass of the coin, radius (r), and square of the angular velocity (ω). |
| 6 | [katex]r = 15 , \text{cm} = 0.15 , \text{m}[/katex] | Convert the radius from centimeters to meters. |
| 7 | [katex]F_{\text{centripetal}} = 0.005 \times 0.15 \times (\frac{60 \times 2 \pi}{60})^2[/katex] | Substitute the values of m, r, and ω into the centripetal force formula. |
| 8 | [katex]F_{\text{centripetal}} , \text{vs} , F_{\text{static max}}[/katex] | Compare the centripetal force with the maximum static friction force to determine if the coin slides off. |
The coin does not slide off the turntable, as the required centripetal force to keep the coin on the turntable is less than the maximum static friction force. The maximum static friction force is approximately 0.03924 N, while the required centripetal force is .0296 N.
Just ask: "Help me solve this problem."
A ball of mass \( m \) is fastened to a string. The ball swings at constant speed in a vertical circle of radius \( R \) with the other end of the string held fixed. Neglecting air resistance, what is the difference between the string’s tension at the bottom of the circle and at the top of the circle?
A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius \(R\). The toy completes each revolution of its motion in a time period \(T\). What is the magnitude of the acceleration of the toy (in terms of \(T\), \(R\), and \(g\))?
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?
A car moving around a circular track with a constant speed has
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
No.
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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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