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

1 | a_t = 1.4 \, \text{m/s}^2 | The given tangential acceleration ( a_t ) of the car is 1.4 \, \text{m/s}^2 . |

2 | a_c = \frac{v^2}{r} | Centripetal acceleration ( a_c ) formula, where v is the velocity of the car and r is the radius of the circular track. |

3 | a_c = a_t | Centripetal acceleration is equal to tangential acceleration as per the question’s condition. |

4 | v = a_t t | Velocity ( v ) as a function of time ( t ) given constant acceleration. This is derived using the formula for velocity under constant acceleration v = u + at , with the initial velocity u = 0 . |

5 | \frac{(a_t t)^2}{r} = a_t | Substituting the expression for v from step 4 into the centripetal acceleration formula from step 2. |

6 | a_t t^2 = r | Solving the equation from step 5, a_t cancels out on both sides. Solving for t^2 . |

7 | t = \sqrt{\frac{r}{a_t}} | Isolating t by taking the square root of both sides of the equation. |

8 | r = \frac{230 \, \text{m}}{2} | Since the diameter of the track is 230 meters, the radius r is half of the diameter. |

9 | t = \sqrt{\frac{115 \, \text{m}}{1.4 \, \text{m/s}^2}} | Plugging the radius r and tangential acceleration a_t into the equation from step 7. |

10 | t \approx 9.06 \, \text{s} | Calculating the actual time when the magnitudes of the centripetal and tangential accelerations are equal, yielding approximately 9.06 seconds. |

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- Statistics

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

Advanced

Mathematical

GQ

A car is moving up the side of a circular roller coaster loop of radius 12 m. The angular velocity is 1.8 \, \text{rad/s} and angular acceleration is -0.82 \, \text{rad/s}^2 . The car is at the same elevation as the center of the loop. Find the magnitude and direction of the acceleration.

- Centripetal Acceleration, Rotational Kinematics

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

Advanced

Mathematical

FRQ

The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.

- Circular Motion, Energy

Advanced

Mathematical

GQ

A 2.2 \times 10^{21} \, \text{kg} moon orbits a distant planet in a circular orbit of radius 1.5 \times 10^8 \, \text{m}. It experiences a 1.1 \times 10^{19} \, \text{N} gravitational pull from the planet. What is the moon’s orbital period in earth days?

- Circular Motion, Gravitation

Intermediate

Mathematical

FRQ

A 2 kg ball is swung in a vertical circle. The length of the string the ball is attached to is 0.7 m. It takes 0.4 s for the ball to travel one revolution ( assume ball travels at constant speed).

- Circular Motion, Tension

Advanced

Mathematical

GQ

Two wires are tied to the 500 g sphere shown below. The sphere revolves in a horizontal circle at a constant speed of 7.2 m/s. What is the tension in the upper wire? What is the tension in the lower wire?

- Circular Motion

Intermediate

Mathematical

GQ

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has coefficients of static and kinetic friction of µ_{s} = 0.80 and µ_{k} = 0.50. The turntable slowly speeds up to 60 rpm. Does the coin slide off the turntable?

- Circular Motion

Intermediate

Conceptual

MCQ

Which of the following do not affect the maximum speed that a car can drive in a circle? Choose both correct answers.

- Circular Motion

Advanced

Mathematical

FRQ

The International Space Station has a mass of 4.2 x10^{5} kg and orbits Earth at a distance of 4.0 x10^{2} km above the surface. Earth has a radius of 6.37 x10^{6} m, and mass of 5.97 x10^{24} kg. Calculate the following:

- Circular Motion, Gravitation

t \approx 9.06 \, \text{s}

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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_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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