The bar AB from Prob 572 is repeated here If the velocity of

The bar AB from Prob. 5/72 is repeated here. If the velocity of point A is 3 m/s to the right and is constant for an interval including the position shown, determine the tangential acceleration of point B along its path and the angular acceleration of the bar.

Solution

Solution:

here it is given bar is moving with constant velocity for interval shown

therefore resultant acceleration acting at bar is zero Ar=0.

As we can see part B of bar is in circular motion, therefore centripetal acceleration towards centre will act due to change in direction of velocity though v is constant.

Ac = v^2/r = 9/0.5

= 18 m/s2

Ar = sqrt(Ac^2 + At^2)

0 = sqrt (18^2 + At^2)

At = 18 m/s2

Angular acceleration is given by = At/(R)

= 18/1.2

= 15 rad/s2