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## Q. 1.4

The slender rod$\left(\overline{I} =\frac{1}{12}mL^{2}\right)$ AC of Figure 1.16(a) of mass m is pinned at B and held horizontally by a cable at C. Determine the angular acceleration of the bar immediately after the cable is cut.

## Verified Solution

Immediately after the cable is cut, the angular velocity is zero. The bar has a fixed axis of rotation at B. Applying Equation (1.35)

$\sum{M_{B} }= \sum{I_{B}\alpha }$      (a)

$\sum{M_{\omicron} } + I_{\omicron }\alpha$                  (1.35)

to the FBD of Figure 1.16(b) and taking moments as positive clockwise, we have

$mg\frac{L}{4}=I_{B}\alpha$                      (b)

The parallel-axis theorem is used to calculate $I_{B}$ as

$I_{B} =\overline{I} +md^{2} =\frac{1}{12}mL^{2}+m\left(\frac{L}{4} \right) ^{2} =\frac{7}{48} mL^{2}$                             (c)

Substituting into Equation (b) and solving for $\alpha$ yields

$\alpha =\frac{12g}{7L}$           (d)

ALTERNATIVE METHOD

Free-body diagrams showing effective and external forces are shown in Figure 1.16(c). The appropriate moment equation is

$\left(\sum{M_{B} } \right) _{ext} =\left(\sum{M_{B} } \right) _{eff}$              (e)

$mg\frac{L}{4}=\frac{1}{12}mL^{2} +\left(m\frac{1}{4} \alpha \right) \left(\frac{L}{4} \right)$               (f)
And $\alpha =\frac{12g}{7L}$