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

 

 

Recall

 ˗  (7/8)m vRxa ˗ (7/8)m vo cos θ  =  (1/3) ML ωa  ˗ (1/3) ML ω2           (4)

 

              e  = ˗ [(7/8L)ωa ˗ vRxa] /[ (7/8L)ω2 ˗ vRxb]                                             (5)

 

and   ˗ vRxb  =  vo cos θ

 

 

Now you have two equations in the two unknowns:   vRxa  and  ωa

 

From eq. (5),  e [ (7/8L)ω2 + vo cos θ]  = vRxa  ˗  (7/8L)ωa     so

 

vRxa  =   (7/8L)ωa  +  e [7/8L)ω2 + vo cos θ]  =   (7/8L)ωa  +  (7/8)e Lω2 + e vo cos θ    (6)

 

Put eq. (6) into eq. (4) and solve for  ωa  .  The result is:

 

ωa  = { [ (1/3)M ˗ (49/64)e m ] L ω2  ˗  (7/8) (1+e) m vo cos θ } / [ (1/3)M + (49/64)m ] L

 

Recall data:  W = 32 oz, m = 5 oz, L = 36 inches, e = 0.8, vo = 88 ft/sec, θ = 30o

 

ωa  =  3.77 rad/sec  so    ωa  =  3.77 k  rad/sec                                (result)

 

Put into eq. (6) to find  vRxa 

 

  vRxa  = 141.2 ft/sec  and recall  vRya = vRyb = vo sin θ   =  44 ft/sec

 

 

Finally express the velocity of the baseball,  vR,  in terms of horizontal and vertical components.           Need a coordinate transformation.

 

                                                              

i = cos θ I  +  sin θ J   and  j = - sin θ I  +  cos θ J   where  θ  =  30o

 

 

vR  =  141.2 i  +  44 j  =  141.2 (cos θ I  +  sin θ ) + 44 (- sin θ I  +  cos θ J )

 

vR  =  100.3 I  +  108.7 J  ft/sec   | vR |  =  147.9 ft/sec                 (result)

 

 

 


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