AP Physics C Mechanics- 5.6 Newton’s Second Law in Rotational Form- Study Notes- New Syllabus
AP Physics C Mechanics- 5.6 Newton’s Second Law in Rotational Form – Study Notes
AP Physics C Mechanics- 5.6 Newton’s Second Law in Rotational Form – Study Notes – per latest Syllabus.
Key Concepts:
- Change in Angular Velocity
- Relationship Between Net Torque, Angular Acceleration, and Rotational Inertia
- Combined Linear and Rotational Analysis
Change in Angular Velocity
The angular velocity of a rigid system changes whenever a net torque is exerted on it. If the torques acting on the system are not balanced (\( \mathrm{\sum \tau \neq 0} \)), the system experiences an angular acceleration (\( \mathrm{\alpha} \)).
Key Idea: A nonzero net torque produces a change in the rate of rotation (i.e., a change in angular velocity), similar to how a nonzero net force produces linear acceleration in translational motion.
Example
A flywheel is acted on by a single torque of \( \mathrm{12\,N·m} \). If the torques are unbalanced, what happens to its rotational motion?
▶️ Answer / Explanation
Since \( \mathrm{\sum \tau \neq 0} \), the flywheel is not in rotational equilibrium. By Newton’s Second Law for rotation, \( \mathrm{\sum \tau = I\alpha} \), so a net torque causes an angular acceleration \( \mathrm{\alpha} \).
Result: The flywheel’s angular velocity changes over time — it speeds up if the net torque acts in the direction of rotation or slows down if it acts in the opposite direction.
Relationship Between Net Torque, Angular Acceleration, and Rotational Inertia
The rate of change of angular velocity, or the angular acceleration, is directly proportional to the net torque exerted on a rigid system and inversely proportional to its rotational inertia:
\( \mathrm{\alpha_{sys} = \dfrac{\sum \tau}{I_{sys}} = \dfrac{\tau_{net}}{I_{sys}}} \)
- \( \mathrm{\alpha_{sys}} \): angular acceleration of the system (\( \mathrm{rad/s^2} \))
- \( \mathrm{\tau_{net}} \): net torque acting on the system (\( \mathrm{N·m} \))
- \( \mathrm{I_{sys}} \): rotational inertia of the system (\( \mathrm{kg·m^2} \))
Key Relationships:
- \( \mathrm{\alpha \propto \tau_{net}} \): Larger torque → larger angular acceleration.
- \( \mathrm{\alpha \propto \dfrac{1}{I}} \): Larger rotational inertia → smaller angular acceleration for the same torque.
- The direction of \( \mathrm{\alpha} \) is the same as the direction of \( \mathrm{\tau_{net}} \), determined by the right-hand rule.
Example
A uniform solid disk of mass \( \mathrm{3.0\,kg} \) and radius \( \mathrm{0.25\,m} \) experiences a net torque of \( \mathrm{0.60\,N·m} \). Find its angular acceleration.
▶️ Answer / Explanation
Step 1: Moment of inertia for a solid disk: \( \mathrm{I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(3.0)(0.25)^2 = 0.0938\,kg·m^2} \).
Step 2: Apply rotational form of Newton’s second law:
\( \mathrm{\alpha = \dfrac{\tau_{net}}{I}} \)
\( \mathrm{\alpha = \dfrac{0.60}{0.0938} = 6.4\,rad/s^2} \)
Result: The disk experiences an angular acceleration of \( \mathrm{6.4\,rad/s^2} \) in the direction of the applied torque.
Combined Linear and Rotational Analysis
To fully describe the motion of a rotating rigid system, both linear and rotational analyses may need to be performed independently and then related through the connection between linear and angular quantities.
Relationships between linear and angular variables:
- \( \mathrm{v = r\omega} \)
- \( \mathrm{a_t = r\alpha} \)
- \( \mathrm{F_{tan} = \tau / r} \)
Thus, if a system both translates and rotates (such as rolling objects), analyzing the motion requires both translational (forces, linear acceleration) and rotational (torques, angular acceleration) dynamics.
Example
A solid cylinder of mass \( \mathrm{2.0\,kg} \) and radius \( \mathrm{0.10\,m} \) rolls without slipping down an incline of angle \( \mathrm{30^\circ} \). Find its linear acceleration.
▶️ Answer / Explanation
Step 1: Use both translational and rotational forms of Newton’s second law.
Translational: \( \mathrm{mg\sin\theta – F_f = ma} \)
Rotational: \( \mathrm{F_f r = I\alpha} \)
and since the cylinder rolls without slipping: \( \mathrm{a = r\alpha} \).
Step 2: Substitute \( \mathrm{I = \tfrac{1}{2}MR^2} \) and eliminate \( \mathrm{F_f} \):
\( \mathrm{mg\sin\theta = Ma + \dfrac{1}{2}Ma} \Rightarrow \mathrm{mg\sin\theta = \tfrac{3}{2}Ma} \)
Step 3: Solve for \( \mathrm{a} \):
\( \mathrm{a = \dfrac{2}{3}g\sin\theta = \dfrac{2}{3}(9.8)\sin(30°) = 3.27\,m/s^2} \)
Result: The cylinder accelerates linearly down the incline at \( \mathrm{3.3\,m/s^2} \). Both linear and rotational analyses were required to solve this.