Sol. The tangential displacement, velocity and acceleration of a point on the rim of pulley 1 or 2 equal the corresponding vertical motions of point A or B since the cables are assumed to be inextensible.

Case : With A momentarily at rest, line AB rotates to AB ′ through the angle d θ during time dt . From the diagram we see that the displacements and their time derivatives gives

ds B = d θ v B = (a B ) t =

ds O = d θ v O = a O =

With v D = r 2 ω 2 = 4(2) = 8 in./sec and a D = r 2 α 2

= 4(–3) = –12 in./sec 2 ,

We have for the angular motion of the double pulley

ω = = = 8/12

= 2/3 rad/sec (CCW) Ans.

α = = = –12/12

= –1 rad/sec 2 (CW) Ans.

The corresponding motion of O and the load L is

v O = = 4(2/3) = 8/3 In./sec Ans.

a O = = 4 (–1) = – 4 in./sec 2 Ans.

Case . With point C, and hence point A, in motion, line AB moves to A ′ B ′ during time dt . From the diagram for this case, we see that the displacements and their time derivatives give

ds B –ds A = d θ v B – v A = (a B ) t –(a A ) t =

ds O –ds A = v O –v A = a O – (a A ) t =

With v C = r 1 ω 1 = 4 (1) = 4 in./sec

v D = r 2 ω 2 = 4(2) = 8 in./sec

a C = r 1 α 1 = 4 (4) = 16 in./sec 2 a D = r 2 α 2 = 4 (–2) = –8 in./sec

2 we have for the angular motion of the double pulley

ω = = =

= 1/3 rad/sec (CCW) Ans.

α = = =

= –2 rad/sec

2 (CW) Ans.

The corresponding motion of O and the load L is

v O = v A + = v C + = 4 + 4 (1/3)

= 16/3 in./sec Ans.

a O = (a A ) t + = a C + = 16 + 4 (–2)

= 8 in./sec 2 Ans.

Helpful Hints :

(i) Recognizer that the inner pulley is a wheel rolling along the fixed line of the left-hand cable. Thus, the expressions of previous problem hold.

(ii) Since B moves along a curved path, in addition to its tangential component of acceleration (a B ) t , it will also have a normal component of acceleration toward O which does not affect the angular acceleration of the pulley

(iii) The diagrams show these quantities and the simplicity of their linear relationships. The visual picture of the motion of O and B as AB rotates through the angle d θ should clarify the analysis.

(iv) Again, as in case , the differential rotation of line AB as seen from the figure establishes the relation between the angular velocity of the pulley and the linear velocities of points A, O and B. The negative sign for (a B ) t = a D produces the acceleration diagram shown but does not destroy the linearity of the relationships.