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

1 (a) | v = r \omega | For rolling motion, the linear velocity v at the bottom is related to the angular velocity \omega by the radius r . Here, the radius r is 0.16 m. |

2 (a) | \omega = \frac{v}{r} | Calculate the angular velocity \omega at the bottom using the given linear velocity v = 3.2 \, \text{m/s} . |

3 (a) | \omega = \frac{3.2 \, \text{m/s}}{0.16 \, \text{m}} = 20 \, \text{rad/s} | Substitute the values into the equation to find \omega . |

4 (a) | \alpha = \frac{\omega}{t} | Angular acceleration \alpha is calculated using the angular velocity \omega and the time t it takes to reach that angular velocity. |

5 (a) | v^2 = v_0^2 + 2aL | Using the kinematic equation for linear motion, where a is linear acceleration and L = 1.5 \, \text{m} is the length of the incline. |

6 (a) | 3.2^2 = 0 + 2a \times 1.5 | a = \frac{(3.2)^2}{2 \times 1.5} = \frac{10.24}{3} \approx 3.413 \, \text{m/s}^2 |

7 (a) | a = r\alpha | Relate linear acceleration a to angular acceleration \alpha . |

8 (a) | \alpha = \frac{a}{r} = \frac{3.413 \, \text{m/s}^2}{0.16 \, \text{m}} \approx 21.33 \, \text{rad/s}^2 | Compute angular acceleration \alpha . |

1 (b) | \omega = \alpha t | Use the equation for angular velocity \omega related to angular acceleration \alpha and time t . |

2 (b) | \omega_{\text{rpm}} = 7329 = \omega \frac{60}{2\pi} | Convert \omega from rpm to rad/s for calculation. |

3 (b) | \omega = 7329 \cdot \frac{2\pi}{60} \approx 767.23 \, \text{rad/s} | Find \omega in rad/s. |

4 (b) | t = \frac{\omega}{\alpha} = \frac{767.23}{419} \approx 1.83 \, \text{s} | Compute the time t . |

1 (c) | \Delta \omega = \alpha \Delta t | Angular velocity change \Delta \omega is given, calculate the time \Delta t using angular acceleration \alpha . |

2 (c) | \Delta \omega = 33.3 \, \text{rad/s} – 3.33 \, \text{rad/s} = 29.97 \, \text{rad/s} | Calculate the change in angular velocity \Delta \omega . |

3 (c) | \Delta t = \frac{\Delta \omega}{\alpha} = \frac{29.97}{5.15} \approx 5.82 \, \text{s} | Calculate the time \Delta t needed to reach the final angular velocity. |

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- Statistics

Advanced

Mathematical

MCQ

A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?

- Rotational Energy, Rotational Inertia, Rotational Kinematics, Rotational Motion, Torque

Beginner

Mathematical

GQ

- Rotational Kinematics

Advanced

Mathematical

GQ

A rotating merry-go-round makes one complete revolution in 4.0 s. What is the linear speed and acceleration of a child seated 1.2 m from the center?

- Rotational Kinematics

Beginner

Mathematical

GQ

A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?

- Rotational Kinematics

Beginner

Mathematical

MCQ

When a fan is turned off, its angular speed decreases from 10 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the average angular acceleration of the fan?

- Rotational Kinematics, Rotational Motion

(a) Angular acceleration of the ball is \approx 21.33 \, \text{rad/s}^2 .

(b) Time for the CD player to reach full speed is \approx 1.83 \, \text{s} .

(c) Time to accelerate for the rotating object is \approx 5.82 \, \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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