Section 1.4: Rotational Dynamics
This section introduces rotational motion concepts, including torque, angular acceleration, moment of inertia, and the rotational analogues of Newton’s laws.
- Torque: \( \tau = r F \sin\theta \)
- Moment of Inertia: \( I = \sum m_i r_i^2 \) or integral \( I = \int r^2 dm \)
- Rotational Newton’s Second Law: \( \tau_{net} = I \alpha \)
- Rotational Kinetic Energy: \( K_{rot} = \frac{1}{2} I \omega^2 \)
- Angular Momentum: \( L = I \omega \), conserved if \( \tau_{ext} = 0 \)
Example 1
A solid disc of mass 2 kg and radius 0.5 m is subjected to a tangential force of 4 N at its rim. Find the angular acceleration.
Moment of inertia for disc: \( I = \frac{1}{2} m R^2 = 0.25 \text{ kg·m²} \)
Angular acceleration: \( \alpha = \frac{\tau}{I} = \frac{R F}{I} = \frac{0.5*4}{0.25} = 8 \text{ rad/s²} \)
Example 2
A thin rod of length 2 m and mass 3 kg rotates about one end under a perpendicular force of 6 N at its tip. Find the angular acceleration.
Moment of inertia: \( I = \frac{1}{3} m L^2 = \frac{1}{3}*3*2^2 = 4 \text{ kg·m²} \)
Torque: \( \tau = F L = 6 * 2 = 12 \text{ N·m} \)
Angular acceleration: \( \alpha = \tau/I = 12/4 = 3 \text{ rad/s²} \)
Practice Problems
- A solid sphere (m = 2 kg, r = 0.3 m) is spun by a tangential force of 5 N at its rim. Find angular acceleration.
- A wheel of radius 0.4 m and mass 4 kg has torque 10 N·m applied. Compute angular acceleration.
- Thin rod of length 1.5 m and mass 2 kg rotated at one end by 3 N force. Find α.
- A disc of radius 0.25 m, mass 1.5 kg, rotates under 6 N·m torque. Find ω after 2 s starting from rest.
- Solid cylinder (m = 3 kg, r = 0.2 m) experiences torque 8 N·m. Find α.
- A hoop of radius 0.5 m and mass 2 kg is spun with τ = 4 N·m. Compute α.
- Thin rod pivoted at center, L = 2 m, m = 3 kg, torque 5 N·m applied. Find α.
- A uniform disc (r = 0.3 m, m = 2 kg) angular acceleration under tangential F = 3 N. Find α.
- Wheel radius 0.2 m, torque 6 N·m, I = 0.15 kg·m². Find α.
- Solid sphere, mass 1 kg, r = 0.2 m, torque 2 N·m. Find α.