The end rollers of bar AB(1.5R) are constrained to the slot. If roller A has a downward velocity of 1.2 m/s and this speed is constant over a small motion interval, determine the tangential acceleration of roller B as it passes the topmost position. The value of R is 0.5 m.

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Lauren Andrews · Accepted Answer

First, we use the Pythagorean theorem in order to calculate the vertical distance (h) between AA and BB. Also, we have established a coordinate system in which i points to the right and textbf{j}j points upwards.  h  =      (    1.5    R          )      2        −          R      2          =  1.1180  R    1  =  1.1180  (  0.5  )  =  0.5590We continue to write the velocity vector at B  (      v    B    )) in terms of the velocity at A   (      v    A    )) and the angular velocity   (  ω  ). Note that       r          A      B       is the position vector from A to B      v    B    =      v    A    +  ω  +      r          A      B          ⇒  −      v    B    i  =  −  1.2  j  +  ω  k  ×  (  0.5  i  +  0.5990  j  )    ⇒  −      v    B    i  =  −  1.2  j  +  0.5  ω  j  −  0.5590  w  iFrom this last expression, we obtain a system of two equations. We solve for omegaω in the second equation. It is important to remark that we had defined   ω as   ω  k      {                                        v            B                    =          −          0.5590          ω                                      0          =          −          1.2          +          0.5          ω                            ⇒  ω  =      1.2    0.5    =  2.4            r      a      d        s      ⇒  ω  =  2.4  k            r      a      d        s  After having obtained   ω, we use the first equation (of the system of equations) in order to calculate       v          B              ​ , we are able to calculate the magnitude of the normal (centripetal) acceleration at B   (      a          B      ,      n        )      v          B        =  0.5590  (  2.4  )  =  1.3416      m    s          a          B      ,      n        =            v      B      2        R      =            (      1.3416              )        2              0.5    =  3.60      m          s      2

The end rollers of bar AB(1.5R) are constrained to the slot. If roller A has a downward velocity of 1.2 m/s and this speed is constant over a small motion interval, determine the tangential acceleration of roller B as it passes the topmost position. The value of R is 0.5 m.

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