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Kinetic Energy of a Body in a Plane

Example: In the figure shown below, bar AB of mass m₁ has an angular velocity of ω₁ k rad/sec,

bar BC of mass m₂ has an angular velocity of ω₂ k rad/sec, and disk, with center C,

of mass M and radius r has an angular velocity of ˗ ω₃ k rad/sec. Let the length of bar AB

be L₁ and bar AB L₂. All pin connections are smooth. D denotes the center of bar AB.

E denotes the center of the bar BC. Calculate the kinetic energy of bar AB and the disk.

Recall that he result the total kinetic energy of the rigid body, T (for plane motion) is

T = ½ m v_C² + ½ I^C_zz ω²

KE of translation + KE of rotation

The kinetic energy for bar AB then equals ½ m₁ v_D² + ½ I^C_zz ω₁² where I^C_zz = (1/12) m₁ L₁²

From kinematics: v_D = v_A + v_D_/A= ω₁ k x (- L₁/2) i = - L₁ ω₁/2 j

T_AB = ½ m₁(- L₁ ω₁/2)² + ½ [(1/12) m₁ L₁²] ω₁² = 1/6 m₁ L₁² ω₁² (result)

Note also that bar AB rotates about the smooth pin at A. So for pure rotation

the kinetic energy for bar AB is

T_AB = ½ I^A_zz ω₁² = ½ (1/3 m₁ L₁²) ω₁² = 1/6 m₁ L₁² ω₁² (same result)

The kinetic energy of the disk is T_disk = ½ m v_C² + ½ I^C_zz ω₃²

The mass moment of inertia of the disk about its mass center, I^C_zz = ½ M r²

From kinematics: v_C = v_B + v_C_/B = v_A + v_B_/A + v_C_/B = 0 + v_B_/A + v_C_/B

v_C = ω₁ k x (- L₁) i + ω₂ k x (- L₂) j = L₂ω₂ i - L₁ ω₁ j

v_C² = v_C ● v_C = ( L₂ω₂ )² + (L₁ ω₁)² So the kinetic energy of the disk is

T_disk = ½ M [( L₂ω₂ )² + (L₁ ω₁)² ] + ½ M r² ω₃² (result)