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Chapter 1

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.

Step-by-Step

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)

leading to

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}