Rotational Dynamics Lab: Moment of Inertia and Torque
Investigate the rotational analogue of mass, the **Moment of Inertia ($I$)**, and its relation to **Torque ($\tau$)** and **Angular Acceleration ($\alpha$)** using the fundamental equation: $\tau = I\alpha$.
Key Equations & Concepts
â–¼$$ I = \Sigma m_i r_i^2 $$ For a Disc: $I = \frac{1}{2}MR^2$. For a Rod (center): $I = \frac{1}{12}ML^2$.
Torque ($\tau$) causes angular acceleration ($\alpha$). $$ \tau = I \cdot \alpha $$ (Units: $\tau$ in $\text{N}\cdot\text{m}$, $I$ in $\text{kg}\cdot\text{m}^2$, $\alpha$ in $\text{rad/s}^2$).
Torque is the product of Force ($F$) and the perpendicular distance ($r$) from the axis: $$ \tau = F \cdot r $$
Experiment 1: Moment of Inertia ($I$) Analyzer
Select a shape and adjust its dimensions to calculate its Moment of Inertia about the central axis.
Experiment 2: Rotational Second Law Challenge ($\tau = I\alpha$)
Find the Angular Acceleration ($\alpha$) given the applied Torque ($\tau$) and Moment of Inertia ($I$).
Moment of Inertia ($I$) is the rotational equivalent of mass ($m$). Torque ($\tau$) is the rotational equivalent of Force ($F$). Rotational motion is fully described by replacing linear variables with angular variables ($F \to \tau$, $m \to I$, $a \to \alpha$).
Theorems and Advanced Concepts (11th Std)
Used to find the Moment of Inertia ($I$) about any axis parallel to the center of mass axis: $$ I = I_{cm} + M d^2 $$
The rotational equivalent of linear momentum ($p$): $$ L = I \omega $$ Conservation of angular momentum ($L=\text{constant}$) is key in phenomena like figure skaters speeding up when pulling in their arms.
In pure rolling motion, the translational kinetic energy and rotational kinetic energy combine: $$ KE_{\text{total}} = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2 $$
