Plane Motion of Rigid Bodies: Energy and Momentum

Why it matters

Understanding the plane motion of rigid bodies using energy and momentum principles is crucial for designing and analyzing mechanical systems such as engines, machinery, and vehicles. These principles help engineers predict the behavior of systems under various forces and motions, ensuring safety and efficiency.

Key ideas

• Plane Motion of Rigid Bodies: Involves the movement of bodies in a two-dimensional plane, where all points in the body move in parallel planes.
• Energy Methods: Utilize the work-energy principle, which states that the work done by all forces acting on a body is equal to the change in its kinetic energy.
• Momentum Methods: Involve the conservation of linear and angular momentum to analyze the motion of rigid bodies.
• Kinetic Energy of Rigid Bodies: Includes translational and rotational components, calculated as the sum of the kinetic energy due to linear motion and the kinetic energy due to rotation.
• Work-Energy Principle: T1 + U1-2 = T2, where T1 and T2 are the initial and final kinetic energies, and U1-2 is the work done by external forces.
• Conservation of Momentum: In the absence of external forces, the total momentum of a system remains constant.

Reference points and conservation conditions

For the planar kinetic-energy formula below, v is the speed of the centre of mass G and I = I_G is the mass moment of inertia about G, perpendicular to the motion plane. Angular momentum H_G = I_G ω uses that same axis. About a fixed point O, the general planar expression is H_O = r_G/O × mv_G + I_G ω; do not omit the translational contribution.

Linear impulse–momentum gives ∫ΣF_ext dt = m(v_G2 − v_G1). Angular impulse–momentum about G gives ∫ΣM_G dt = I_G(ω₂ − ω₁) for planar motion. Linear momentum is conserved when the net external impulse is zero; angular momentum about G is conserved when the net external angular impulse about G is zero. These are separate conditions.

External work includes work of forces at their moving application points and work of couples, ∫M dθ. A fixed support may exert a force but do no work.

Formulas

• T = 1/2·m·v^2 + 1/2·I·ω^2
  • T: Total kinetic energy (Joules)
  • m: Mass of the body (kg)
  • v: Linear velocity (m/s)
  • I: Moment of inertia (kg·m²)
  • ω: Angular velocity (rad/s)
• L = I·ω
  • L: Angular momentum (kg·m²/s)
  • I: Moment of inertia (kg·m²)
  • ω: Angular velocity (rad/s)

Worked example

Given: A rigid body with mass m = 10 kg, moment of inertia I = 2 kg·m², initial linear velocity v1 = 3 m/s, and initial angular velocity ω1 = 2 rad/s. The net work of all external forces and couples is U1-2 = 50 J. Here v1 is the centre-of-mass speed and I is I_G.

1. Calculate initial kinetic energy (T1):

   • T1 = 1/2·m·v1^2 + 1/2·I·ω1^2
   • T1 = 1/2·10·(3)^2 + 1/2·2·(2)^2
   • T1 = 45 + 4 = 49 J

2. Apply work-energy principle to find final kinetic energy (T2):

   • T1 + U1-2 = T2
   • 49 + 50 = T2
   • T2 = 99 J

3. Check whether the final velocities are determined:

   • Assume v2 and ω2 such that T2 = 1/2·m·v2^2 + 1/2·I·ω2^2
   • This requires solving for v2 and ω2 based on additional conditions or constraints.

Final Answer: T2 = 99 J

Common mistakes

• Confusing linear and angular quantities, such as velocity and momentum.
• Neglecting the rotational component of kinetic energy.
• Incorrectly applying the work-energy principle by not accounting for all forces.

For GATE ME

Questions often involve calculating the kinetic energy of a system, applying the work-energy principle, or using conservation of momentum. Practice problems that require analyzing both translational and rotational motion.

Quick check

1. What is the work-energy principle?
2. How is angular momentum calculated for a rigid body?
3. What are the components of kinetic energy in plane motion?

Answers: 1. The work done by forces is equal to the change in kinetic energy. 2. For planar motion about the centre of mass, H_G = I_G·ω. 3. Translational and rotational kinetic energy.