^{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?

Step | Formula / Calculation | Reasoning |
---|---|---|

1 | m = 2.00 \times 10^2 , \text{g} = 0.200 , \text{kg} | Convert mass from grams to kilograms. |

2 | r = 50.0 , \text{cm} = 0.500 , \text{m} | Convert length of string from centimeters to meters. |

3 | \text{rpm} = 75.0 , \text{rpm} | Given revolutions per minute. |

4 | \omega = \text{rpm} \times \frac{2\pi}{60} | Convert rpm to radians per second. |

5 | \omega = 75.0 \times \frac{2\pi}{60} | Substituting the given rpm value. |

6 | v = \omega \times r | Formula for linear speed in circular motion. |

7 | v = 3.93 , \text{m/s} | Calculating the speed of the block. |

8 | T = \frac{mv^2}{r} | Tension in the string formula. |

9 | T = 6.17 , \text{N} | Calculating the tension in the string. |

The speed of the block is approximately 3.93 m/s and the tension in the string is approximately 6.17 N.

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- Statistics

Intermediate

Conceptual

MCQ

A delivery truck is traveling north. It then turns along a leftward circular curve. This the packages in the truck to slide to the RIGHT. Which of the following is true of the net force on the packages as they are sliding?

- 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

Advanced

Proportional Analysis

MCQ

A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius *R* and is vertical. In terms of *R *and constants, find the minimum height *h *above the bottom of the loop the block must start from so it makes it around the loop.

- Circular Motion

Intermediate

Mathematical

FRQ

A 1.00 kg mass is attached to a 0.800 m long string and spun in a vertical circle. The mass completes 2.00 revolution in 1.00 s.

- Circular Motion

Advanced

Conceptual

MCQ

A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?

- Angular Momentum, Kepler's Law, Rotational Energy

Intermediate

Conceptual

MCQ

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

- Circular Motion, Gravitation

Advanced

Proportional Analysis

MCQ

Two identical object rests on a platform rotating at constant speed. Object A is at distance of half the platform’s radius from the center. Object B lays at edge of the platform. Assuming the platform continues rotating at the same speed, how does the centripetal force of the two objects compare?

- Circular Motion

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

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

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

3.93 m/s, 6.17 N

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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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