Question 17.13: Determine the angular acceleration of the spool in Fig. 17-2......

SOLUTION I

Free-Body Diagram. Fig. 17-20b. The 100-N force causes a_{G} to act upward. Also, a acts clockwise, since the spool winds around the cord at A.

There are three unknowns T, a_{G} , and α. The moment of inertia of the spool about its mass center is

I_{G} = m k²_{G} = 8  kg(0.35 m)²= 0.980   kg · m²

Equations of Motion.

+ ↑  Σ  F_{y}  = m(a_{G})_{y};        T   +   100   N   –   78.48   N =  (8   kg) a_{G}                   (1)

↻+    ΣM_{G}  =   I_{G}α ;    100   N(0.2   m)   –   T(0.5   m) =  (0.980   kg · m²)α                        (2)

Kinematics. A complete solution is obtained if kinematics is used to
relate a_{G} to α. In this case the spool “rolls without slipping” on the cord at A. Hence, we can use the results of Example 16.4 or 16.15, so that

(↻+) a_{G} = αr .            a_{G} =     α(0.5  m)                 (3)

Solving Eqs. 1 to 3, we have

α      = 10.3 rad/s²

a_{G}  = 5.16   m/s²

T = 19.8   N

SOLUTION II

Equations of Motion. We can eliminate the unknown T by summing
moments about point A. From the free-body and kinetic diagrams
Figs. 17-20b and 17-20c, we have

↻+    ΣM_{A}  = Σ( M_{k})_{A} ;   100  N(0.7  m)   –   78.48  N(0.5  m) 

= (0.980   kg · m²)α   +   [(8   kg)a_{G}](0.5   m)

Using Eq. (3),

α  = 10.3   rad/s²

SOLUTION III

Equations of Motion. The simplest way to solve this problem is to
realize that point A is the IC for the spool. Then Eq. 17-19 applies.

ΣM_{IC}  = I_{IC}α          17-19

↻+    ΣM_{A}  = I_{A}α ;   ; (100   N) (0.7   m)  –   (78.48   N)(0.5  m)

= [0.980  kg .m² + (8  kg) (0.5  m)²]α

α = 10.3  rad/s²