Lecture Notes: Parallel Axis Theorem and Moment of Inertia

Introduction

• Topic: Parallel Axis Theorem
• System: Solid Sphere and Rod
• Objective: Calculate the moment of inertia for different axes of rotation.

Key Concepts

• Moment of Inertia: Depends on mass distribution relative to the axis of rotation.
• Parallel Axis Theorem:
  • Formula: ( I = I_{cm} + md^2 )
  • ( I_{cm} ) = Moment of inertia through center of mass
  • ( m ) = Mass
  • ( d ) = Distance between center of mass and axis of rotation

Case 1: Rotation about a Distant Axis

• Configuration: Sphere and rod rotating about an axis outside the sphere.
• Sphere:
  • ( I_{sphere} = \frac{2}{5} m_s r^2 + m_s (l + r)^2 )
  • Given: ( m_s = 10 ) kg, ( r = 0.5 ) m, ( l = 2 ) m
  • Calculation: ( I_{sphere} = 63.5 ) kg·m²
• Rod:
  • Formula: ( I_{rod} = \frac{1}{3} ml^2 )
  • Given: ( m = 5 ) kg
  • Calculation: ( I_{rod} = 6.67 ) kg·m²
• Total Moment of Inertia: ( 70.2 ) kg·m²

Case 2: Rotation at Edge of Sphere

• Configuration: Axis of rotation at sphere’s edge.
• Sphere:
  • ( I_{sphere} = \frac{7}{5} m_s r^2 )
  • Calculation: ( I_{sphere} = 3.5 ) kg·m²
• Rod:
  • Same as Case 1: ( I_{rod} = 6.67 ) kg·m²
• Total Moment of Inertia: ( 10.17 ) kg·m²

Case 3: Rotation Through Center of Sphere

• Configuration: Axis of rotation passes through the center of the sphere.
• Sphere:
  • ( I_{sphere} = \frac{2}{5} m_s r^2 )
  • Calculation: ( I_{sphere} = 1 ) kg·m²
• Rod:
  • Formula: ( I = I_{cm} + m(d^2) )
  • ( d = \frac{l}{2} + r )
  • Calculation: ( I_{rod} = 12.92 ) kg·m²
• Total Moment of Inertia: ( 13.92 ) kg·m²

Observations and Conclusions

• Greatest Moment of Inertia: When mass distribution is farthest from the axis.
• Smallest Moment of Inertia: When axis passes through the center of mass.

Additional Notes

• The lecture demonstrated how the Parallel Axis Theorem can be used to calculate the moment of inertia for complex systems.
• For additional questions, comments can be made under the lecture video.
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