Rotational Motion and Moment of Inertia - Quick Notes

Rotational Motion Fundamentals

Newton's second law for rotation: torque is related to angular acceleration by \tau = I \alpha. For a single force at radius R, the torque is \tau = F R, and the tangential acceleration is a_t = \alpha R. For a point mass at distance R, the moment of inertia is I = m R^2, so the torque relation becomes F R = m \alpha R^2.

Moment of Inertia

Moment of inertia is the rotational analogue of mass: I = \sumi mi r_i^2 or I = \int r^2 \; dm. For common shapes:

• Rod about its center: I = \frac{1}{12} M L^2
• Rod about end: I = \frac{1}{3} M L^2
• Solid cylinder (solid disk) about central axis: I = \frac{1}{2} M R^2
• Solid sphere: I = \frac{2}{5} M R^2
• Hollow cylinder: I = M R^2
• Hollow sphere: I = \frac{2}{3} M R^2

Angular Dynamics

Torque and angular acceleration: T = I \alpha. If a force F acts at a radius R, the torque is T = F R and with no slipping the linear-to-angular relation is a = \alpha R. For a system with a single rotating mass, this mirrors Newton's second law in linear form with constant mass and radius.

Rotational Energy and Mechanical Energy

Rotational kinetic energy: K{\text{rot}} = \frac{1}{2} I \omega^2; Translational kinetic energy: K{\text{trans}} = \frac{1}{2} m v^2. Mechanical energy: E = PE + K{\text{trans}} + K{\text{rot}}.

Rolling without slipping and Energy

When rolling without slipping v = \omega R and a = \alpha R. For a body rolling down a height h without slipping, the bottom speed satisfies
v^2 = \frac{2 g h}{1 + I/(m R^2)}.
For common shapes, the ratio I/(m R^2) values.

Rolling down Inclines: Sphere vs Cylinder

Two objects with same mass and radius; sphere has I/(m R^2) = 2/5 and cylinder I/(m R^2) = 1/2. Therefore the sphere attains a larger bottom speed and reaches the bottom first:

• Sphere: v^2 = \frac{2 g h}{1 + 2/5} = \frac{10}{7} g h
• Cylinder: v^2 = \frac{2 g h}{1 + 1/2} = \frac{4}{3} g h

Angular Momentum

Angular momentum is L = I \omega. In the absence of external torque, angular momentum is conserved: Li = Lf \\Rightarrow Ii \omegai = If \omegaf. A common example: a person on a frictionless stool holds weights; when weights are pulled inward, the system's moment of inertia decreases and the angular speed increases to conserve L.

Work and Energy in Rotation

Work done on a rotating object changes its rotational kinetic energy: W = \Delta K{rot} = \frac{1}{2} If \omegaf^2 - \frac{1}{2} Ii \omegai^2. In problems with combined translation and rotation, one uses the total kinetic energy K{tot} = K{trans} + K{rot} and W = \Delta K_{tot}.

Example: Reels and Buckets (Torque and Acceleration)

For a reel of mass M and radius R with a hanging bucket of mass m:

• Moment of inertia of reel: I = \frac{1}{2} M R^2 (solid cylinder)
• Equations: m g - T = m a, and T R = I \alpha, with a = \alpha R
• Solution: a = \dfrac{m g}{m + I/R^2}. Numerically, with m=2\,\text{kg}, M=3\,\text{kg}, R=0.4\,\text{m}, g=9.8\,\text{m/s}^2: I = 0.24\,\text{kg m}^2, I/R^2 = 1.5, a \approx 5.6\,\text{m/s}^2 downward for the bucket.

Angular Momentum Problem: Inertia Change on Stool

Given initial inertia Ii and angular speed \omegai, and final inertia If after weights move inward, angular momentum conservation gives Ii \omegai = If \omegaf \\Rightarrow \omegaf = \dfrac{Ii}{If} \omegai. The example values yield \omegaf \approx 6.55\,\text{rad/s} when Ii = 2.25 and If = 1.08 with \omega_i \approx 5.24\,\text{rad/s}.

Rotational Work Example

If an object speeds up from \omegai to \omegaf with inertia changing from Ii to If, the work done equals
W = \Delta K{rot} = \frac{1}{2} If \omegaf^2 - \frac{1}{2} Ii \omega_i^2
For a sample calculation, the work was found to be about 7.72 \text{ J} in the given setup.