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 v on a curve of radius R is F_c = \frac{mv^2}{R}. The component of the normal force providing this centripetal force on a banked curve without friction is F_{c0} = \frac{mv_0^2}{R}.

The additional centripetal force required due to the increased speed must be provided by static friction f_s, which is \mu_s N, where N is the normal force and \mu_s is the coefficient of static friction.

Step | Formula / Derivation | Reasoning |
---|---|---|

1 | F_{c0} = \frac{mv_0^2}{R} | Centripetal force for design speed. |

2 | F_{c} = \frac{mv^2}{R} | Centripetal force for actual speed. |

3 | f_s = F_{c} – F_{c0} | Additional force required from static friction. |

4 | f_s = \mu_s N | Static friction force. |

5 | N = mg \cos(\theta) | Normal force on banked curve, where \theta is the banking angle. |

6 | \tan(\theta) = \frac{v_0^2}{Rg} | Relationship for a banked curve without friction. |

7 | \mu_s = \frac{f_s}{N} | Coefficient of static friction. |

8 | \mu_s = \frac{F_{c} – F_{c0}}{mg \cos(\theta)} | Combine steps 3, 4, and 5. |

9 | \mu_s = \frac{\frac{mv^2}{R} – \frac{mv_0^2}{R}}{mg \cos(\theta)} | Substitute F_{c} and F_{c0} from steps 1 and 2. |

10 | \mu_s = \frac{v^2 – v_0^2}{Rg \cos(\theta)} | Mass m cancels out. |

11 | \mu_s = \frac{53^2 – 40^2}{125 \cdot 9.8 \cdot \cos(\theta)} | Substitute values for v, v_0, and R. |

12 | \mu_s = \frac{2809 – 1600}{1225 \cdot 9.8 \cdot \cos(\theta)} | Square the speeds. |

13 | \mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \cos(\theta)} | Subtract the squares. |

14 | \mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \frac{40^2}{125 \cdot 9.8}} | Substitute \cos(\theta) from step 6. |

15 | \mu_s = \frac{1209 \cdot 125}{1225 \cdot 40^2} | Simplify. |

16 | \boxed{\mu_s = \frac{1209}{1960}} | Final calculation for \mu_s. |

The coefficient of static friction required is approximately \frac{1209}{1960}, which can be simplified further if needed.

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- Statistics

Intermediate

Mathematical

GQ

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 6 \times 10^{24} \text{ kg} the the radius of Earth to be 6.4 \times 10^6 \text{ m}. What is the satellite’s velocity?

- Centripetal Acceleration, Circular Motion, Gravitation, Linear Forces

Beginner

Mathematical

GQ

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*)?

- Circular Motion

Advanced

Mathematical

MCQ

- Circular Motion

Advanced

Mathematical

GQ

An Olympic bobsled team goes through a horizontal curve at a speed of 120 km/hr. If the radius of curvature is 10.0 m, what is the apparent weight the crew experiences-express in terms of mg.

- Circular Motion

Intermediate

Mathematical

GQ

A 2.00 x10^{2} g block on a 50.0 cm long string swings in a circle on a horizontal, frictionless table at 75.0 rpm. What is the speed of block? What is the tension in the string?

- Circular Motion

\frac{1209}{1960} \approx .61

Kinematics | Forces |
---|---|

\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |

v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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