Energy and Momentum of Rotating Systems · ⁨พลังงานและโมเมนตัมของระบบหมุน⁩

Learning Objective 6.1.A: Describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system.

• 6.1.A.1 The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system and is given by the equation $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
  • Equation: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$
  • 6.1.A.1.i The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy.
  • 6.1.A.1.ii The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass.
• 6.1.A.2 A rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.
• 6.1.A.3 Rotational kinetic energy is a scalar quantity.

Learning Objective 6.2.A: Describe the work done on a rigid system by a given torque or collection of torques.

• 6.2.A.1 A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.
• 6.2.A.2 The amount of work done on a rigid system by a torque is related to the magnitude of that torque and the angular displacement through which the rigid system rotates during the interval in which that torque is exerted.
  • Equation: $W = \displaystyle\int_{\theta_1}^{\theta_2} \tau \, d\theta$
• 6.2.A.3 Work done on a rigid system by a given torque can be found from the area under the curve of a graph of the torque as a function of angular position.

Learning Objective 6.3.A: Describe the angular momentum of an object or rigid system.

• 6.3.A.1 The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$.
  • Equation: $L = I\omega$
• 6.3.A.2 The angular momentum of an object about a given point is $\vec{L} = \vec{r} \times \vec{p}$.
  • Equation: $\vec{L} = \vec{r} \times \vec{p}$
  • 6.3.A.2.i The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.
  • 6.3.A.2.ii The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.

Learning Objective 6.3.B: Describe the angular impulse delivered to an object or rigid system by a torque.

• 6.3.B.1 Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted.
  • Equation: $\text{angular impulse} = \displaystyle\int \tau \, dt$
• 6.3.B.2 Angular impulse has the same direction as the torque imparting it.
• 6.3.B.3 The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.

Learning Objective 6.3.C: Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system.

• 6.3.C.1 The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial momenta of the object or rigid system.
  • Equation: $\Delta L = L - L_0$
• 6.3.C.2 A rotational form of the impulse–momentum theorem relates the angular impulse delivered to an object or rigid system and the change in angular momentum of that object or rigid system.
  • 6.3.C.2.i The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
    • Equation: $\Delta L = \displaystyle\int_{t_1}^{t_2} \tau \, dt$
  • 6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant.
    • Equation: $\tau_{\text{net}} = \dfrac{dL}{dt} = I\dfrac{d\omega}{dt} = I\alpha$
• 6.3.C.3 The net torque exerted on an object or rigid system is equal to the slope of the graph of the angular momentum of an object as a function of time.
• 6.3.C.4 The angular impulse delivered to an object or rigid system is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.

Learning Objective 6.4.A: Describe the behavior of a system using conservation of angular momentum.

• 6.4.A.1 The total angular momentum of a system about a rotational axis is the sum of the angular momenta of the system's constituent parts about that rotational axis.
• 6.4.A.2 Any change to a system's angular momentum must be due to an interaction between the system and its surroundings.
  • 6.4.A.2.i The angular impulse exerted by one object or system on a second object or system is equal and opposite to the angular impulse exerted by the second object or system on the first. This is a direct result of Newton's third law.
  • 6.4.A.2.ii A system may be selected so that the total angular momentum of that system is constant.
  • 6.4.A.2.iii The angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or farther from the rotational axis.
  • 6.4.A.2.iv If the total angular momentum of a system changes, that change will be equivalent to the angular impulse exerted on the system.

Learning Objective 6.4.B: Describe how the selection of a system determines whether the angular momentum of that system changes.

• 6.4.B.1 Angular momentum is conserved in all interactions.
• 6.4.B.2 If the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant.
• 6.4.B.3 If the net external torque exerted on a selected object or rigid system is nonzero, angular momentum is transferred between the system and the environment.

Learning Objective 6.5.A: Describe the kinetic energy of a system that has translational and rotational motion.

• 6.5.A.1 The total kinetic energy of a system is the sum of the system's translational and rotational kinetic energies.
  • Equation: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$

Learning Objective 6.5.B: Describe the motion of a system that is rolling without slipping.

• 6.5.B.1 While rolling without slipping, the translational motion of a system's center of mass is related to the rotational motion of the system itself with the following equations:
  • Equation: $\Delta x_{\text{cm}} = r\Delta\theta$
  • Equation: $v_{\text{cm}} = r\omega$
  • Equation: $a_{\text{cm}} = r\alpha$
• 6.5.B.2 For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.

Learning Objective 6.5.C: Describe the motion of a system that is rolling while slipping.

• 6.5.C.1 When slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related.
• 6.5.C.2 When a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system.

Boundary statement: Rolling friction is beyond the scope of AP Physics C: Mechanics.

Learning Objective 6.6.A: Describe the motions of a system consisting of two objects or systems interacting only via gravitational forces.

• 6.6.A.1 In a system consisting only of a massive central object and an orbiting satellite with mass that is negligible in comparison to the central object's mass, the motion of the central object itself is negligible.
• 6.6.A.2 The motion of satellites in orbits is constrained by conservation laws.
  • 6.6.A.2.i In circular orbits, the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are constant.
  • 6.6.A.2.ii In elliptical orbits, the system's total mechanical energy and the satellite's angular momentum are constant, but the system's gravitational potential energy and the satellite's kinetic energy can each change.
  • 6.6.A.2.iii The gravitational potential energy of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object.
    • Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
• 6.6.A.3 The total energy of a system consisting of a satellite orbiting a central object in a circular path can be written in terms of the gravitational potential energy of that system or the kinetic energy of the satellite.
  • Equation: $K = -\dfrac{1}{2}U$
  • Equation: $E_{total} = \dfrac{1}{2}U = -\dfrac{GMm}{2r}$
• 6.6.A.4 The escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the satellite–central-object system is equal to zero.
  • 6.6.A.4.i When the only force exerted on a satellite is gravity from a central object, a satellite that reaches escape velocity will move away from the central body until its speed reaches zero at an infinite distance from the central body.
  • 6.6.A.4.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
    • Equation: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$