Angular Impulse/Momentum for Rigid Bodies in a Plane (example continued)

Impulsive forces act at the handle of the bat, O, are ∫ O_y dt and ∫O_x dt likewise the impulsive force between the baseball and the bar is ∫F dt as shown in the figure below.

Note: The impulsive force between the bat and the baseball are equal in magnitude and

opposite indirection. Let C denote the center of mass of the bat located at L/2 along the bat.

Let E denote the location where the baseball strikes the bat. Let the subscript a denote the

condition immediately after impact and the subscript b denote the condition immediately

before impact. “b” is the same as position 2 determined using the principle of work and energy.

Apply the principle of linear impulse and momentum to the bat. The result is:

∫ O_x dt i + ∫ O_y dt j - ∫ F dt i = M (v_Cxa – v_Cxb) i + M (v_Cya – v_Cyb) j

From the x-component for linear momentum:

∫ O_x dt - ∫ F dt = M (v_Cxa – v_Cxb) (1)

Apply the principle of angular impulse and momentum to the bat. The result is:

( - L/2) ∫ O_x dt k - (3L/8) ∫ F dt k = (1/12) (ML²) (ω_a – ω_b) k or in scalar form

( - L/2) ∫ O_x dt - (3L/8) ∫ F dt = (1/12) (ML²) (ω_a – ω₂) (2)

Multiply eq. (1) by L/2 and add to eq. (2)

- (7L/8) ∫ F dt = M(L/2)(v_Cxa – v_Cxb) + (1/12) (ML²) (ω_a – ω₂) (3)

Use kinematics to relate linear and angular motion. i.e.

v_Cxa = (L/2) ω_a , v_Cxb = (L/2) ω₂ and put into (3). Click here to continue with this example.

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