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: $I = \sum<em>i m</em>i r_i^2$ 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: $K<em>{\text{rot}} = \frac{1}{2} I \omega^2$; Translational kinetic energy: $K</em>{\text{trans}} = \frac{1}{2} m v^2$. Mechanical energy: $E = PE + K<em>{\text{trans}} + K</em>{\text{rot}}$.

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: $L<em>i = L</em>f \\Rightarrow I<em>i \omega</em>i = I<em>f \omega</em>f$. 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: $W = \Delta K<em>{rot} = \frac{1}{2} I</em>f \omega<em>f^2 - \frac{1}{2} I</em>i \omega<em>i^2$. In problems with combined translation and rotation, one uses the total kinetic energy $K</em>{tot} = K<em>{trans} + K</em>{rot}$ 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 $I<em>i$ and angular speed $\omega</em>i$, and final inertia $I<em>f$ after weights move inward, angular momentum conservation gives $I</em>i \omega<em>i = I</em>f \omega<em>f \\Rightarrow \omega</em>f = \dfrac{I<em>i}{I</em>f} \omega<em>i.$ The example values yield $\omega</em>f \approx 6.55\,\text{rad/s}$ when $I<em>i = 2.25$ and $I</em>f = 1.08$ with $\omega_i \approx 5.24\,\text{rad/s}$.

Rotational Work Example

If an object speeds up from $\omega<em>i$ to $\omega</em>f$ with inertia changing from $I<em>i$ to $I</em>f$, the work done equals
$W = \Delta K<em>{rot} = \frac{1}{2} I</em>f \omega<em>f^2 - \frac{1}{2} I</em>i \omega_i^2$
For a sample calculation, the work was found to be about $7.72 \text{ J}$ in the given setup.