AP Physics 1- 5.6 Newton’s Second Law in Rotational Form- Study Notes- New Syllabus
AP Physics 1-5.6 Newton’s Second Law in Rotational Form – Study Notes
AP Physics 1-5.6 Newton’s Second Law in Rotational Form – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Newton’s Second Law in Rotational Form
Newton’s Second Law in Rotational Form
Newton’s second law for rotation relates the net external torque acting on a rigid body to its angular acceleration. It is the rotational analog of Newton’s second law of linear motion.
\( \sum \tau = I \alpha \)
Where:
- \( \tau \) = Torque (Nm)
- \( I \) = Moment of inertia (kg·m²)
- \( \alpha \) = Angular acceleration (rad/s²)
Key Points:
- Just as \( F = ma \) governs linear motion, \( \tau = I \alpha \) governs rotational motion.
- The greater the rotational inertia \( I \), the harder it is to change the angular motion of the system.
- When torque is zero, angular acceleration is zero, meaning the object rotates with constant angular velocity.
- When torque is nonzero, the object experiences angular acceleration proportional to the net torque.
Relation with Linear Form:
Linear: \( F = m a \) ↔ Rotational: \( \tau = I \alpha \)
Graphical Representation:
- The slope of an angular velocity vs. time graph gives angular acceleration \( \alpha \).
- The slope of an angular momentum vs. time graph gives the net torque \( \tau \).
Conditions for Change in Angular Velocity:
- A system’s angular velocity changes only when a net external torque acts on it.
- If \( \sum \tau \neq 0 \), then \( \alpha \neq 0 \), meaning angular velocity changes with time.
- If \( \sum \tau = 0 \), then \( \alpha = 0 \), and the system rotates with constant angular velocity (or remains at rest).
Example :
A solid disk of mass 4 kg and radius 0.5 m is acted upon by a tangential force of 10 N at its rim. Find the angular acceleration of the disk. (For a solid disk, \( I = \dfrac{1}{2} M R^2 \))
▶️ Answer/Explanation
Torque: \( \tau = F \cdot R = 10 \times 0.5 = 5 \, \text{Nm} \)
Moment of inertia: \( I = \dfrac{1}{2} (4)(0.5^2) = 0.5 \, \text{kg·m}^2 \)
Angular acceleration: \( \alpha = \dfrac{\tau}{I} = \dfrac{5}{0.5} = 10 \, \text{rad/s}^2 \)
Answer: \( \alpha = 10 \, \text{rad/s}^2 \)
Example :
A wheel with a moment of inertia \( 2 \, \text{kg·m}^2 \) experiences a constant torque of \( 8 \, \text{Nm} \). Find its angular acceleration.
▶️ Answer/Explanation
From Newton’s second law: \( \alpha = \dfrac{\tau}{I} = \dfrac{8}{2} = 4 \, \text{rad/s}^2 \)
Answer: \( \alpha = 4 \, \text{rad/s}^2 \)
Example :
A uniform rod of length 2 m and mass 6 kg is pivoted at one end. A force of 20 N is applied perpendicularly at the free end. Find the angular acceleration of the rod. (For a rod pivoted at one end, \( I = \dfrac{1}{3} M L^2 \))
▶️ Answer/Explanation
Torque: \( \tau = F \cdot L = 20 \times 2 = 40 \, \text{Nm} \)
Moment of inertia: \( I = \dfrac{1}{3} (6)(2^2) = \dfrac{1}{3} \times 24 = 8 \, \text{kg·m}^2 \)
Angular acceleration: \( \alpha = \dfrac{\tau}{I} = \dfrac{40}{8} = 5 \, \text{rad/s}^2 \)
Answer: \( \alpha = 5 \, \text{rad/s}^2 \)