Section 5.2: Rotational Dynamics
This section introduces the causes of rotational motion: torques and moment of inertia. We explore how forces generate angular acceleration.
Key Concepts:
- Torque: \( \tau = r F \sin\theta \)
- Moment of inertia: \( I = \sum m_i r_i^2 \) (or integral form for continuous objects)
- Newton’s 2nd Law for rotation: \( \tau_\text{net} = I \alpha \)
Example 1
A 2 kg mass is attached 0.5 m from the rotation axis. A force of 10 N is applied perpendicular. Find angular acceleration.
\( \tau = r F = 0.5 \times 10 = 5 \, \text{N·m} \)
\( I = m r^2 = 2 \times 0.5^2 = 0.5 \, \text{kg·m²} \)
\( \alpha = \tau / I = 5 / 0.5 = 10 \, \text{rad/s²} \)
Example 2
A uniform rod of length 2 m and mass 3 kg is pivoted at one end. Find the torque required for angular acceleration α = 4 rad/s² at center of mass.
Moment of inertia about pivot: \( I = \frac{1}{3} M L^2 = 1/3 * 3 * 2^2 = 4 \, \text{kg·m²} \)
Torque: \( \tau = I \alpha = 4 * 4 = 16 \, \text{N·m} \)
Practice Problems
- A 5 kg mass on a 1 m radius disk. Force 15 N applied tangentially. Find α.
- Uniform rod 1.5 m, mass 2 kg, pivot at end. Find torque for α = 3 rad/s².
- Wheel of I = 0.8 kg·m², τ_net = 4 N·m. Find angular acceleration.
- Torque of 10 N·m acts on disk I = 2 kg·m². Find α and angular speed after 5 s starting from rest.
- Force applied at 0.4 m from axis of rotation 8 N. Find α for mass 1.5 kg at that radius.