The wheel is moving to the right such that it has an angular

The wheel is moving to the right such that it has an angular velocity omega = 1.9 rad/s and angular acceleration a = 4.8 rad/s2 at the instant shown. 1.If it does not slip at A, determine the magnitude of the acceleration of point B. 2.Determine the direction of the acceleration of point B.

Solution

given alpha = 4.8rad/sec²   and w= 1.9 rad/sec

since no slipping

ac= alpha*r= 4.8*1.45= 6.96 ft/sec²

a_B= ac + alpha*r_B/C -w² r_B/C

a_B= 6.96i+(-4.8k)*(-1.45cos30i+1.45sin30j)-1.9² (-1.45cos30i+1.45sin30j)

=6.96i-3.61(-1.45*0.86i+1.45*0.5j)

=6.96i+4.5i-2.6j=11.46i-2.6j

a_B=sqrt (131.33+6.76)=sqrt (138.09)=11.75ft/sec²