Sol.

If the cable does not slip on, the drum, the vertical velocity and acceleration of the load L are, of necessity, the same as the tangential velocity v and tangential acceleration a t of point C. For the rectilinear motion of L with constant acceleration, the n- and t-components of the acceleration of C become

[v 2 = 2as] a=a t = v 2 /2s = 3 2 /[2(4))]=1.125 ft/sec 2 [a n = v

2 /r] a n = 3 2 /(24/12) = 4.5 ft/sec 2 [a = ] a c =

= 4.64 ft/sec

2 . Ans.

The angular motion of gear A is determined from the angular motion of gear B by the velocity v 1 and tangential acceleration a 1 of their common point of contact. First, the angular motion of gear B is determined from the motion of point C on the attached drum. Thus,

[v = r ω ] ω B = v/r = 3/(24/12) = 1.5 rad/sec

[a t = r α ] α B = α t /r= 1.125/(24/12) =0.562 rad/sec 2 Then from v 1 = r A ω A = r B ω B and a 1 = r A α A

= r B α B , we have

ω A = ω B = 1.5 = 4.5 rad/sec CW Ans.

α A = α B = 0.562 = 1.688 rad/sec

2 CW Ans.

Helpful Hint:

Recognize that a point on the cable changes the direction of its velocity after it contacts the drum and acquires a normal component of acceleration.