Question 21.2: The rod in Fig. 21-9a has a weight per unit length of 1.5 lb......

The principle of impulse and momentum will be used since impact occurs.

Impulse and Momentum Diagrams. Fig. 21-9b. During the short time Δt, the impulsive force F acting at A changes the momentum of the rod. (The impulse created by the rod’s weight W during this time is small compared to \int{} \textbf{F}  dt, so that it can be neglected, i.e., the weight is a non impulsive force.) Hence, the angular momentum of the rod is conserved about point A since the moment of \int{} \textbf{F} dt about A is zero.

Conservation of Angular Momentum. Equation 21-9 must be used to find the angular momentum of the rod, since A does not become a fixed point until after the impulsive interaction with the hook. Thus, with reference to Fig. 21-9b, (\textbf{H}_{A})_{1} = (\textbf{H}_{A})_{2} or

\textbf{H}_{A} = ρ_{G/A} ×  m\textbf{v}_{G}+ \textbf{H}_{G}             (21-9)

r_{G/A} × m (\textbf{v}_{G} )_{1} = r_{G/A} × m(\textbf{v}_{G} )_{2}   +  (\textbf{H}_{G})_{2}        (1)

From Fig. 21-9a, r_{G/A} = {-0.667i   +   0.5j} ft. Furthermore, the primed axes are principal axes of inertia for the rod because I_{x'y'} = I_{x'z'} = I_{z'y'} = 0. Hence, from Eqs. 21-11, (H_{G})_{2} = I_{x'}\textbf{ω}_{x}\textbf{i} +   I_{y'}\textbf{ω}_{y}\textbf{j}+ I_{z'}\textbf{ω}_{z}\textbf{k}. The principal moments of inertia are I_{x'} = 0.0272  slug· ft², I_{y'} = 0.0155  slug· ft², I_{z'} = 0.0427  slug· ft² (see Prob. 21-13). Substituting into Eq. 1, we have

H_{x} = I_{x}\textbf{ω}_{x}          H_{y} = I_{y} \textbf{ω}_{y}           H_{z} = I_{z} \textbf{ω}_{z}               (21-11)

(-0.667\textbf{i}+   0.5\textbf{j})  ×  [(\frac{4.5}{32.2} ) (-10\textbf{k})] = (-0.667\textbf{i} + 0.5\textbf{j}) × [(\frac{4.5}{32.2} ) (-  \textbf{v}_{G} )_{2} \textbf{k}]

+ 0.0272\textbf{ω}_{x}\textbf{i} +   0.0155\textbf{ω}_{y}\textbf{j} +  0.0427\textbf{ω}_{z}\textbf{k}

Expanding and equating the respective i, j, k components yields

–  0.699 = –  0.0699 (\textbf{v}_{G} )_{2}  + 0.0272\textbf{ω}_{x}                  (2)

– 0.932 = – 0.0932(\textbf{v}_{G} )_{2}  +   0.0155\textbf{ω}_{y}                     (3)

o = 0.0427\textbf{ω}_{z}                                                                        (4)

Kinematics. There are four unknowns in the above equations; however, another equation may be obtained by relating w to (\textbf{v}_{G} )_{2} using kinematics. Since \textbf{ω}_{z} = 0 (Eq. 4) and after impact the rod rotates about the fixed point A, Eq. 20-3 can be applied, in which case (\textbf{v}_{G} )_{2} = w × r_{G/A},or

– (\textbf{v}_{G} )_{2} \textbf{k}= (\textbf{ω}_{x}\textbf{i} +  \textbf{ω}_{y}\textbf{j})  ×   (- 0.667\textbf{i} +   0.5\textbf{j})

– (\textbf{v}_{G} )_{2} = 0.5\textbf{ω}_{x} +   0.667\textbf{ω}_{y}                   (5)

Solving Eqs. 2, 3 and 5 simultaneously yields

(\textbf{v}_{G} )_{2} = {-8.41k} ft/s                ω = {-4.09i   –   9.55j} rad/s