(a) Calculate the linear speed of the sphere when it reaches the bottom of the incline.

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

1 | h = L \sin(\theta) | Calculate the vertical height h fallen by the sphere using the length of the incline L and the sine of the incline angle \theta. |

2 | h = 7.0 \sin(35^\circ) | Substitute L = 7.0 \, m and \theta = 35^\circ. |

3 | PE_{\text{top}} = KE_{\text{trans}} + KE_{\text{rot}} | Use the conservation of mechanical energy, where potential energy at the top is equal to the sum of transnational kinetic energy and rotational kinetic energy at the bottom. |

4 | mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 | Express the conservation of energy equation in terms of v (linear velocity) and \omega (angular velocity). |

5 | I = \frac{2}{5}MR^2 | Substitute the given moment of inertia for a solid sphere, where I = \frac{2}{5}MR^2. |

6 | v = R\omega | Relation between linear velocity and angular velocity for rolling without slipping. |

7 | \omega = \frac{v}{R} | Rearrange the equation for \omega. |

8 | mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 | Substitute I and \omega in terms of v and R. |

9 | mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 | Simplify the equation by canceling M and R. |

10 | mgh = \frac{7}{10}mv^2 | Combine like terms. |

11 | v^2 = \frac{10}{7}gh | Isolate v^2. |

12 | v = \sqrt{\frac{10}{7}gh} | Take the square root to find v. |

13 | v = \sqrt{\frac{10}{7}(9.8)(7.0 \sin(35^\circ))} | Substitute the values of g and h. |

14 | \boxed{v \approx 7.5 \, \text{m/s}} | Final answer. |

(b) Determine the angular speed of the sphere at the bottom of the incline.

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

1 | \omega = \frac{v}{R} | Use the relation between linear and angular velocities for a rolling object. |

2 | \omega = \frac{7.5}{0.15} | Substitute v = 7.5 \, \text{m/s} and R = 0.15 \, m (converted from cm). |

3 | \boxed{\omega \approx 50 \, \text{rad/s}} | Final answer. |

(c) Does the linear speed depend on the radius or mass of the sphere?

Step | Analysis | Conclusion |
---|---|---|

1 | From the energy conservation equation, the mass m cancelled out and the final expression for v didn’t include the radius R. | The linear speed does not depend on the mass or radius of the sphere as both factors were eliminated in deriving v. |

(d) Does the angular speed depend on the radius or mass of the sphere?

Step | Analysis | Conclusion |
---|---|---|

1 | Angular speed \omega was found from v divided by R, but it did not involve mass m. | Angular speed does depend on the radius and does not depend on the mass of the sphere. |

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- Statistics

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

Mathematical

MCQ

The moment of inertia of a solid cylinder about its axis is given by I = \frac{1}{2}mR^2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

- Rotational Energy, Rotational Motion

Intermediate

Mathematical

MCQ

A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.

- Angular Momentum, Momentum, Rotational Motion, Torque

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

Intermediate

Mathematical

GQ

In lacrosse, a typical throw is made by rotating the stick through an angle of roughly 90°, then releasing the ball when the stick is vertical, as shown above. If the 1 meter long stick is at rest when horizontal and the ball leaves the stick with a velocity of 10 m/s, what angular acceleration must the stick experience?

- Rotational Kinematics

- v \approx 7.5 \, \text{m/s}
- \omega \approx 50 \, \text{rad/s}
- According to the equation derived in part (a) the speed does
**not**depend on the radius or mass of sphere. - Angular speed
**does**depend on the radius and does**not**depend on the mass of the sphere.

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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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