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

A1 | T – Mg = \frac{Mv^2}{L} | This is the force equilibrium equation at the bottom point. T is the tension in the string, M is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and L is the length of the string. |

A2 | T = 3Mg | The tension at the bottom is given to be three times the weight of the ball. |

A3 | 3Mg – Mg = \frac{Mv^2}{L} | Substituting the tension value into the equilibrium equation. |

A4 | v = \sqrt{2gL} | Simplifying the equation for v , observe M cancels out. This is velocity at any given point around the circle. |

A5 | F_{\text{centripetal}} = \frac{Mv^2}{L} | At the top, we can use the velocity calculated in the previous step to find the centripetal force required to keep the ball moving in the circle. |

A6 | 2mg | Substitute in the equation for velocity (from step A4) so that the final equation is in terms of M \, g \, L |

B1 | v_{\text{top}} = \sqrt{v^2 – 4gL} | Using conservation of energy. KE_{\text{bottom}} + PE_{\text{bottom}} = KE_{\text{top}} + PE_{\text{top}} . The velocity at the top is found by noting the potential energy difference between top and bottom. Simplify by substituting v from A4. |

B2 | v_{\text{top}} = \sqrt{2gL – 4gL} | = \sqrt{-2gL} \rightarrow which is zero since 2gL > 4gL |

C1 | t = \sqrt{\frac{4L}{g}} | Ball falls freely under gravity and has no initial vertical velocity, so \delta y = \frac{1}{2}gt^2 ; solving for t gives the time to fall a distance L . Note that the displacement from the top to the bottom is twice the radius of the circle or 2L . |

D1 | \Delta x = v_0t | The horiztontal distance traveled by any projectile is the product of the horiztonal speed and the time in air. |

D2 | \sqrt{2gL} \times \sqrt{\frac{4L}{g}} | Substitute in velocity from Step A4 and time from step C1 |

D3 | \sqrt{8}L | Simplify |

(a) The net force on the ball at the top is 2Mg , downward.

(b) The velocity of the ball at the top is v = \sqrt{2gL} .

(c) The time it takes to reach the ground is \sqrt{\frac{4L}{g}} .

(d) The horizontal distance traveled is \sqrt{8}L

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

- Statistics

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

Intermediate

Mathematical

GQ

An 80 kg person sits in a swing that goes around in a circle. The chain connecting the swing to the center of the ride is 8 m long and it makes and angle of 40° with the horizontal. What is the speed of the person going around in a circle?

- Circular Motion

Advanced

Mathematical

GQ

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?

- Projectiles

Advanced

Mathematical

FRQ

One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance h above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance x . The block is released and strikes the floor a horizontal distance D from the edge of the table. Air resistance is negligible.

Derive an expressions for the following quantities only in terms of M, x, D, h, and any constants.

- Energy, Projectiles

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

(a) 2Mg , downward.

(b) v = \sqrt{2gL} .

(c) \sqrt{\frac{4L}{g}} .

(d) \sqrt{8}L

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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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