A4-3. Rigid Body Mechanics: Dynamics & EnergyHL ONLY

1. Moment of Inertia & Newton's Second Law

In linear mechanics, mass ($m$) represents an object's resistance to linear acceleration. In rotational mechanics, Moment of Inertia ($I$) represents an object's resistance to angular acceleration. Crucially, it depends not just on the total mass, but on how that mass is distributed relative to the axis of rotation. Mass concentrated further from the axis creates a much larger moment of inertia.

Moment of Inertia for discrete point masses:

$$I = \sum m r^2$$

$I$: Moment of inertia in kg m²
$m$: Mass of the point mass (kg)
$r$: Perpendicular distance from the axis of rotation (m)

Newton's Second Law for Rotation:

Just as a net force causes linear acceleration ($F = ma$), a net torque causes angular acceleration. The resulting angular acceleration is directly proportional to the net torque and inversely proportional to the moment of inertia.

$$\Sigma \tau = I \alpha$$

2. Angular Momentum & Angular Impulse

Angular momentum is a measure of the amount of rotation an object has, taking into account its mass, shape, and speed.

Angular Momentum ($L$): The rotational equivalent of linear momentum ($p = mv$).

$$L = I \omega$$

Newton's Second Law (Momentum Form): Just as force is the rate of change of linear momentum ($F = \frac{\Delta p}{\Delta t}$), torque is the rate of change of angular momentum.

$$\Sigma \tau = \frac{\Delta L}{\Delta t}$$

Conservation of Angular Momentum: If no external net torque acts on a closed system ($\Sigma \tau = 0$), the total angular momentum of the system remains perfectly constant ($I_i \omega_i = I_f \omega_f$).
Example: When an ice skater pulls their arms in, their radius decreases, meaning their moment of inertia ($I$) decreases. To conserve angular momentum ($L$), their angular velocity ($\omega$) must proportionately increase, causing them to spin faster.

Angular Impulse ($\Delta L$): The change in angular momentum resulting from a torque applied over a period of time.

$$\text{Angular Impulse } = \tau \Delta t = \Delta L = I\omega_f - I\omega_i$$

3. Rotational Work & Power

Just as a force doing work transfers energy in linear motion, a torque doing work transfers energy in rotational motion.

Work Done by a Torque ($W$): The rotational equivalent of $W = F s$.

$$W = \tau \theta$$

Rotational Power ($P$): The rate at which rotational work is done, equivalent to $P = F v$.

$$P = \tau \omega$$

4. Rotational Kinetic Energy

A rolling object possesses both translational kinetic energy (from its center of mass moving forward) and rotational kinetic energy (from its mass spinning around its axis).

Rotational Kinetic Energy ($E_{k,rot}$):

$$E_{k(rot)} = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$$

Total Kinetic Energy of a Rolling Body:

$$E_{total} = E_{k(trans)} + E_{k(rot)} = \frac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$$

Crucial Condition (Rolling Without Slipping): If an object rolls without skidding, its linear velocity and angular velocity are locked together by the equation $\mathbf{v = \omega r}$. This allows you to substitute $\omega = \dfrac{v}{r}$ into the energy equation to solve complex slope problems.

Example 1: Stacked Flywheels

Problem: A flywheel of mass $M$ and radius $R$ rotates at a constant angular velocity $\omega$ about an axis through its centre. The rotational kinetic energy is $E_k$. The moment of inertia of this flywheel is $I_1 = \dfrac{1}{2}MR^2$.

A second flywheel of mass $\dfrac{1}{2}M$ and radius $\dfrac{1}{2}R$ is placed directly on top of the first. The new angular velocity of the combined flywheels becomes $\dfrac{2}{3}\omega$. What is the new rotational kinetic energy of the combined system in terms of $E_k$?

Solution:

1. State the initial Kinetic Energy ($E_k$): $$ E_k = \frac{1}{2}I_1\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega^2 = \mathbf{\frac{1}{4}MR^2\omega^2} $$
2. Find the total Moment of Inertia ($I_{new}$):
The moment of inertia of the second (smaller) flywheel is:
$$ I_2 = \frac{1}{2}(\frac{1}{2}M)(\frac{1}{2}R)^2 = \frac{1}{2}(\frac{1}{2}M)(\frac{1}{4}R^2) = \frac{1}{16}MR^2 $$ Thus, $$ I_{new} = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{16}MR^2 = \frac{8}{16}MR^2 + \frac{1}{16}MR^2 = \mathbf{\frac{9}{16}MR^2} $$
3. Calculate the new Kinetic Energy ($E_{new}$): $$ \begin{aligned} E_{new} &= \frac{1}{2} I_{new} (\omega_{new})^2 = \frac{1}{2} (\frac{9}{16}MR^2) (\frac{2}{3}\omega)^2 = E_{new} = \frac{1}{2} (\frac{9}{16}MR^2) (\frac{4}{9}\omega^2) \\ &= \frac{1}{2} (\frac{4}{16}) MR^2\omega^2 = \frac{1}{8} MR^2\omega^2 \end{aligned} $$ Since $E_k = \dfrac{1}{4}MR^2\omega^2$, we can see that $E_{new}$ is exactly half of that.
Final Answer: $E_{new} = \dfrac{E_k}{2}$