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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

AP Physics 1-Concise Summary Notes- All Topics

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: \( \alpha = 5 \, \text{rad/s}^2 \)

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