Rotational Dynamics, Energy, and the Parallel Axis Theorem

Rotational Dynamics and Energy

Work-Energy Theorem in Rotational Systems

Rotational motion involves the interplay between forces, torques, and energy. The work-energy theorem can be applied to systems with rotational components to determine final velocities and angular speeds.

• Work-Energy Theorem: The net work done on a system is equal to the change in its kinetic energy.

• Rotational Kinetic Energy: $KE_{rot} = \frac{1}{2} I \omega^2$, where I is the moment of inertia and ω is the angular velocity.

• Example: For a cylinder of mass M and radius R pulled by a force F over a distance d (with no friction):

  • Work done: $W = F \cdot d$

  • Moment of inertia for a solid cylinder: $I = \frac{1}{2}MR^2$

  • Final angular velocity: $\omega = 2 \sqrt{\frac{Fd}{MR^2}}$

  • Final linear velocity: $v = \omega R$

Parallel Axis Theorem

Calculating Moment of Inertia About Different Axes

The parallel axis theorem allows calculation of the moment of inertia about any axis parallel to one passing through the center of mass (CM).

• Theorem Statement: $I = I_{CM} + Md^2$, where d is the distance between axes.

• Application Example: For a thin rod of length L and mass M:

  • Moment of inertia about center: $I_{CM} = \frac{1}{12}ML^2$

  • Moment of inertia about end: $I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$

Energy Conservation in Rotational Systems

Combining Translational and Rotational Energy

When objects rotate and translate, both forms of kinetic energy must be considered. Conservation of energy is a powerful tool for solving such problems.

• Energy Conservation Equation: $KE + U = \text{constant}$

• Example: A mass m falls from height h attached to a rotating drum of mass M:

  • Initial energy: $E_{tot} = mgh$

  • Final energy: $E_{tot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

  • Moment of inertia for drum: $I = \frac{1}{2}MR^2$

  • Relating $v$ and $\omega$: $v = \omega R$

  • Solving for final speed: $v = 2 \sqrt{\frac{gh}{2 + \frac{M}{m}}}$

• Special Case: If $M \ll m$, $v \approx \sqrt{2gh}$ (purely translational drop).

Rotational Kinetic Energy of Spherical Shells

Calculating Energy After Angular Displacement

For rotating shells, the moment of inertia and angular displacement determine the final kinetic energy.

• Moment of Inertia for Spherical Shell: $I = \frac{2}{3}MR^2$

• Angular Displacement: $\Delta \theta$ (in radians)

• Final Angular Velocity: $\omega^2 = \omega_0^2 + 2\alpha \Delta \theta$

• Kinetic Energy After Rotation: $KE = \frac{1}{2}I\omega^2$

• Example: For $M = 820$ kg, $R = 0.220$ m, initial $\omega_0 = 0.870$ rad/s, after $\Delta \theta = 12.57$ rad:

  • Final $\omega = 2 \times 0.870 = 1.74$ rad/s

  • Final $KE = \frac{2}{3}MR^2 \Delta \theta$

Summary Table: Moments of Inertia

The following table summarizes moments of inertia for common objects and axes:

Object	Axis	Moment of Inertia (I)
Solid Cylinder	Central axis	$\frac{1}{2}MR^2$
Thin Rod	Center	$\frac{1}{12}ML^2$
Thin Rod	End	$\frac{1}{3}ML^2$
Spherical Shell	Central axis	$\frac{2}{3}MR^2$

Additional info: These notes expand on the handwritten examples by providing full derivations, definitions, and context for the formulas and theorems used in rotational dynamics and energy conservation.