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

 

 

    Recall eq. (3)   ˗ (7L/8) ∫ F dt   =  M(L/2)(vCxa ˗ vCxb)   +  (1/12) (ML2) (ωa ˗ ω2)          (3)

 

 

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

 

     vCxa = (L/2) ωa ,    vCxb =  (L/2) ω2   and put into (3)  and divide by L

 

˗  (7/8) ∫ F dt   =  ½ ML[( ½ ωa ˗ ½ ω2 )]   +  (1/12) (ML) (ωa ˗  ω2)          or

 

    ˗  (7/8) ∫ F dt   = (1/3) ML( ωa  ˗ ω2 )  

 

 

Next apply the principle of linear impulse and momentum to the baseball.

 

                                               

 

                                  

                            ∫ F dt   =  m( vRxa ˗ vRxb)     and     m( vRya ˗ vRyb)  =  0     or

 

                            ∫ F dt   =  m( vRxa + vo cos θ)     and     m( vRya ˗ vo sin θ)  =  0    

 

 

So        (7/8)∫ F dt   =  (7/8)mvRxa + (7/8)m vo cos θ)     Therefore

 

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

 

Eq. (4) contains two unknowns:     vRxa  and  ωa    So you need more information.

 

 

During impact the relative velocity after impact relates to that before impact by the

coefficient of restitution.

                                                  e  =  ˗ (vExa ˗ vRxa) / (vExb ˗ vRxb)

 

Next use kinematics to express  vExa and vExb  in terms of  ωa and ωb = ω2

 

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

 

Click here to continue with this example.

 

 

 


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