Friction - -     Rigid Bodies in Translation and Rotation  (continued)

 

 

                                              

 

 

Next examine case 2  where the disk rolls with slipping in the + x - direction.

 

          ΣF =  m acx               ˗ Px  ˗ F  + Dx  =  m   aCx                          (1)

 

          ΣF =  m acy =  0         N ˗ mg  + Py  +  Dy  =  0                         (2)  for data  N  =  19 N

 

         ΣMC =  Ic α               ˗ F r  + Dy r  =  (1/2) m r2  α                     (3)

 

And since slipping occurs    F  =  Fmax  =  μ N  =  4.75 N.

 

 

For the given data:

 

By  (1)      aCx   =   [  ˗ Px  ˗ μ N + Dx ] / m  =  ˗ 1.1875 m/sec2 

 

By (3)          α  =   2 [  ˗ μ N  + Dy ] / m r    =  1.90625  rad/sec2    

 

 

Next calculate  aP  .     aP  =  aC  +  aP/C |n  +  aP/C |t  

 

Note:  aP/C |n   =  0  since disk starts from rest  i.e.  ω = 0

 

                  aP  = ˗ 1.1875 i    +  α k  x ˗ r j  =   ˗ 1.1875 i    +  1.90625 k  x ˗ 4 j  = 6.44 i   m/sec2

 

Result:  Since aP  and  F  are in opposite directions, slipping does occur in the + x-direction.

 

Therefore:   aC  =  ˗ 1.1875 i  m/sec2   and   α  =    1.90625 k  rad/sec2    

 

As a check, click here to consider case 3 where the disk rolls and slips in the  ˗ x - direction.

 

Note the free body diagram will need to be revised to show the correct direction for the

assumed friction force, F.

 

                               



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