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For the banked curve, the centripetal force necessary to make the turn is provided by the horizontal component of the normal force. When a car travels faster than the design speed, additional centripetal force must be provided by friction.
The centripetal force required for a car traveling at speed [katex]v[/katex] on a curve of radius [katex]R[/katex] is [katex]F_c = \frac{mv^2}{R}[/katex]. The component of the normal force providing this centripetal force on a banked curve without friction is [katex]F_{c0} = \frac{mv_0^2}{R}[/katex].
The additional centripetal force required due to the increased speed must be provided by static friction [katex]f_s[/katex], which is [katex]\mu_s N[/katex], where [katex]N[/katex] is the normal force and [katex]\mu_s[/katex] is the coefficient of static friction.
| Step | Formula / Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F_{c0} = \frac{mv_0^2}{R}[/katex] | Centripetal force for design speed. |
| 2 | [katex]F_{c} = \frac{mv^2}{R}[/katex] | Centripetal force for actual speed. |
| 3 | [katex]f_s = F_{c} – F_{c0}[/katex] | Additional force required from static friction. |
| 4 | [katex]f_s = \mu_s N[/katex] | Static friction force. |
| 5 | [katex]N = mg \cos(\theta)[/katex] | Normal force on banked curve, where [katex]\theta[/katex] is the banking angle. |
| 6 | [katex]\tan(\theta) = \frac{v_0^2}{Rg}[/katex] | Relationship for a banked curve without friction. |
| 7 | [katex]\mu_s = \frac{f_s}{N}[/katex] | Coefficient of static friction. |
| 8 | [katex]\mu_s = \frac{F_{c} – F_{c0}}{mg \cos(\theta)}[/katex] | Combine steps 3, 4, and 5. |
| 9 | [katex]\mu_s = \frac{\frac{mv^2}{R} – \frac{mv_0^2}{R}}{mg \cos(\theta)}[/katex] | Substitute [katex]F_{c}[/katex] and [katex]F_{c0}[/katex] from steps 1 and 2. |
| 10 | [katex]\mu_s = \frac{v^2 – v_0^2}{Rg \cos(\theta)}[/katex] | Mass [katex]m[/katex] cancels out. |
| 11 | [katex]\mu_s = \frac{53^2 – 40^2}{125 \cdot 9.8 \cdot \cos(\theta)}[/katex] | Substitute values for [katex]v[/katex], [katex]v_0[/katex], and [katex]R[/katex]. |
| 12 | [katex]\mu_s = \frac{2809 – 1600}{1225 \cdot 9.8 \cdot \cos(\theta)}[/katex] | Square the speeds. |
| 13 | [katex]\mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \cos(\theta)}[/katex] | Subtract the squares. |
| 14 | [katex]\mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \frac{40^2}{125 \cdot 9.8}}[/katex] | Substitute [katex]\cos(\theta)[/katex] from step 6. |
| 15 | [katex]\mu_s = \frac{1209 \cdot 125}{1225 \cdot 40^2}[/katex] | Simplify. |
| 16 | [katex]\boxed{\mu_s = \frac{1209}{1960}}[/katex] | Final calculation for [katex]\mu_s[/katex]. |
The coefficient of static friction required is approximately [katex]\frac{1209}{1960}[/katex], which can be simplified further if needed.
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A \(2 \, \text{kg}\) ball is swung in a vertical circle. The length of the string the ball is attached to is \(0.7 \, \text{m}\). It takes \(0.4 \, \text{s}\) for the ball to travel one revolution (assume the ball travels at constant speed).
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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 race car travels in a circular track of radius \( 200 \) \( \text{m} \). If the car moves with a constant speed of \( 80 \) \( \text{m/s} \),

An adult exerts a horizontal force on a swing that is suspended by a rope of length \( L \), holding it at an angle \( \theta \) with the vertical. The child in the swing has a weight \( W \) and dimensions that are negligible. In terms of \( W \) and \( \theta \), determine:
[katex]\frac{1209}{1960} \approx .61 [/katex]
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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