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

1 | \tau = FR | The torque (\tau) exerted on the cylinder is due to the force F applied at a radius R. The formula for torque is the force times the perpendicular distance (radius in this case) from the axis of rotation. |

2 | \tau = I\alpha | Newton’s second law for rotation states that the torque is equal to the moment of inertia (I) times the angular acceleration (\alpha). |

3 | I = \frac{1}{2}MR^2 | The moment of inertia for a solid cylinder about its axis is given by this formula, where M is the mass and R is the radius of the cylinder. |

4 | FR = \frac{1}{2}MR^2 \alpha | Substitute the moment of inertia of the cylinder into the torque equation. |

5 | \alpha = \frac{2F}{MR} | Solve for the angular acceleration (\alpha) by isolating \alpha on one side of the equation. |

6 | \omega^2 = \omega_0^2 + 2\alpha \theta | Use the kinematic equation for rotational motion to relate the angular displacement (\theta) to the final angular velocity (\omega). Here, \omega_0 (initial angular velocity) is zero as the cylinder starts from rest. |

7 | \omega^2 = 2\alpha \theta | Substitute \omega_0 = 0 into the equation because the cylinder starts from rest. |

8 | \omega = \sqrt{2\alpha \theta} = \sqrt{\frac{4F\theta}{MR}} | Substitute the value of \alpha from Step 5 into the equation to find \omega. |

9 | K = \frac{1}{2}I\omega^2 | The total kinetic energy (K) of the rotating cylinder is given by the formula for rotational kinetic energy, where I is the moment of inertia and \omega is the angular velocity. |

10 | K = \frac{1}{2} \times \frac{1}{2}MR^2 \times \left(\frac{4F\theta}{MR}\right) | Substitute the expressions for I and \omega into the kinetic energy formula. |

11 | K = \frac{F\theta R}{2} | Simplify the equation to get the final expression for the kinetic energy. |

12 | K = \frac{F\theta R}{2} |
Conclude with the neat, simplified expression for the kinetic energy of the cylinder after it has rotated through an angle \theta. |

Phy can also check your working. Just snap a picture!

- Statistics

Intermediate

Mathematical

MCQ

A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting?

- Torque

Advanced

Mathematical

MCQ

A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s^{2} on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 1.00 N·m? Assume that the moment of inertia of a rectangle is I = \frac{1}{12}M(a^2+b^2)

- Rotational Inertia, Rotational Motion, Torque

Intermediate

Conceptual

MCQ

Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance d. Ignore friction and assume that an equal force *F* is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?

- Energy

Intermediate

Proportional Analysis

MCQ

If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?

- Energy, Momentum

Advanced

Mathematical

FRQ

A solid sphere of mass 1.5 \, \text{kg} and radius 15 \, \text{cm} rolls without slipping down a 35^\circ incline that is 7 \, \text{m} long. Assume it started from rest. The moment of inertia of a sphere is I= \frac{2}{5}MR^2 .

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

**K = \frac{F\theta R}{2}**

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

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!

The most advanced version of Phy. Currently 50% off, for early supporters.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages
- Unlimited Image Uploads
- Unlimited Smart Actions
- Unlimited UBQ Credits
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- 200% Memory Boost
- 150% Better than GPT
- 75% More Accurate, 50% Faster
- Mobile Snaps
- Focus Mode
- No Ads

A quick explanation

UBQ credits are specifically used to grade your FRQs and GQs.

You can still view questions and see answers without credits.

Submitting an answer counts as 1 attempt.

Seeing answer or explanation counts as a failed attempt.

Lastly, check your average score, across every attempt, in the top left.

MCQs are 1 point each. GQs are 1 point. FRQs will state points for each part.

Phy can give partial credit for GQs & FRQs.

Phy sees everything.

It customizes responses, explanations, and feedback based on what you struggle with. Try your best on every question!

Understand you mistakes quicker.

For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.

Aim to increase your understadning and average score with every attempt!

10 Free Credits To Get You Started

*Phy Pro members get unlimited credits