_{s}) for a car not to skid on the same curve when traveling at 53 m/s?

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 2.2 kg ball on the end of a 0.35 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 5.3 m/s. Find the tension in the string.

- Circular Motion

Beginner

Mathematical

GQ

A car travels at a constant speed around a circular track whose radius is 2.6 km. the car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?

- Circular Motion

Intermediate

Mathematical

FRQ

A linear spring of negligible mass requires a force of 18.0 N to cause its length to increase by 1.0 cm. A sphere of mass 75.0 g is then attached to one end of the spring. The distance between the center of the sphere M and the other end P of the un-stretched spring is 25.0 cm. Then the sphere begins rotating at constant speed in a horizontal circle around the center P. The distance P and M increases to 26.5 cm.

- Circular Motion, 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

Beginner

Mathematical

GQ

A speed skater goes around a turn that has a radius of 31 m. The skater has a speed of 14 m/s and experiences a centripetal force of 460 N. What is the mass of the skater?

- Circular Motion

Advanced

Proportional Analysis

MCQ

A car is safely negotiating an unbanked circular turn at a speed of 17 m/s on dry road. However, a long wet patch in the road appears and decreases the maximum static frictional force to one-fifth of its dry-road value. If the car is to continue safely around the curve, by what factor would the it need to change the original velocity?

- 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

Intermediate

Conceptual

MCQ

A compressed spring mounted on a disk can project a small ball. When the disk is not rotating, as shown in the top view above, the ball moves radially outward. The disk then rotates in a counterclockwise direction as seen from above, and the ball is projected outward at the instant the disk is in the position shown above. Which of the following best shows the subsequent path of the ball relative to the ground?

- Circular Motion

Advanced

Mathematical

GQ

The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample.

If the samples are 12 cm from the axis of the rotor and the ultracentrifuge spins at 60000 rpm, what is the magnitude of the net force on the rotor due to the unbalanced samples?

- Circular Motion

Intermediate

Conceptual

MCQ

Which of the following best explains why astronauts experience weightlessness while orbiting the earth?

- Circular Motion, Gravitation

\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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