In-depth Notes on Torque, Angular Momentum, and Moment of Inertia

Torque, Angular Momentum, and Moment of Inertia Review

Moment of Inertia of a Rod Spinning About its Midpoint

  • Derivation: To find the moment of inertia \( I \) of a rod rotating about its midpoint:

    1. Consider the linear mass density \( ext{λ} = rac{M}{L}</li>
    2. By dividing the rod into infinitesimally small elements: \( ext{dm} = ext{λ} \, ext{dx} </li>
    3. Distance from the pivot to each element at position \( x \) results in \( I = ext{fr}^2 \, dm = ext{∫} rac{M}{L} \, x^2 \, dx </li>
    4. The limits of integration are from -( \frac{L}{2} ) to ( \frac{L}{2} ) giving:

      I = \frac{1}{12} M L^2<br />
    • Conclusion: Hence proved (\text{I }= \frac{1}{12} m l^2).

Moment of Inertia of a Rod Spinning About One End

  • Derivation: For a rod spinning about one of its ends, the calculation changes slightly:
    1. Changing the limits of integration from 0 to L.
    2. Using the same linear mass density approach.
    3. The integration yields:

      I = \frac{1}{3} M L^2<br />
    • Conclusion: Hence proved (\text{I }= \frac{1}{3} m l^2).

Moment of Inertia of a Uniform Disk Spinning About its Center

  • Derivation: For a thin disk:
    1. Area mass density (\text{σ} = \frac{M}{A}), where (A = \pi r^2).
    2. The infinitesimal area is (\text{dA} = 2 \pi r \text{dr}).
    3. Integrating to find moment of inertia:

      I = \int_0^R r^2 2 \pi r \text{dr} = \frac{1}{2} M R^2<br />
    • Conclusion: Hence proved (\text{I }= \frac{1}{2} m r^2).

Rotational Inertia of a Cone About the Pivot

  • Derivation: A cone behaves similar to a rod, with its linear mass density given as (\text{λ} = \beta y^2), yielding:

    I = \int0^{h0} y^2 \lambda dy = \int0^{h0} (\beta y^2) dy\ = \frac{\beta h_0^4}{4}\n
  • Conclusion: Resulting in the necessary inertia about the pivot.

Conservation of Angular Momentum with a Rotating Stool

  • When a person pulls weights towards their chest, the moment of inertia decreases from (5.33 kg·m^2) to (1.60 kg·m^2), leading to conservation of angular momentum:

    I1 \omega1 = I2 \omega2<br />
  • Solving for (\omega2): Given (\omega1 = 3.72 rad/s) leads to an increased angular speed when arms are pulled in.

Practical Applications: Ceiling Fan and Fishing Reel Problems

  1. Ceiling Fan Calculation: Given an angular speed of (2.75 rad/s) and a torque of (0.120 N·m), find the moment of inertia (I): use (\tau = I\alpha).
  2. Fishing Reel Analysis: Pulling with a force of (2.2 N) gives angular acceleration and line pulled calculations.

    \alpha = \frac{T}{I}<br /> \text{where } I = \frac{1}{2} m r^2<br />

Collision and Angular Momentum Transfer in Merry-Go-Round Example

  1. A person of mass 59.4kg jumps on a merry-go-round of mass 155kg and radius 2.63m.
  2. Initial angular momentum is conserved to find final angular speed after collision:
    • Use the conservation laws to integrate momenta.

General Formulas and Relationships

  • General Equations:
    • (L = r \times p = I \omega\r)
    • Angular momentum conservation: (L{initial} = L{final})
    • Moment of inertia for various shapes and conditions can be derived using calculus.

Key Concepts and Formulas

  • Linear mass density: (\text{λ} = \frac{M}{L} )
  • Moment of Inertia: \( ext{I} = \int r^2 dm)
  • Angular momentum: (L = mv), and for rotation: (L = I \omega)

Summary of Calculations

  • Use conservation principles as necessary for rotation-related problems, ensuring dimensional consistency and application of relevant equations.
  • Continuous review of solid shapes’ moments of inertia is crucial for applications in physics and engineering studies.