Discussion
Two identical rigid slender bars of length 2L and mass \(m\) are acted upon by a transverse force F at one of the ends as shown in the figure. In the first case, the bar is pinned at the other end. In the second case, the bar is pinned at its mid-point. What should be the magnitude of the force F such that the resulting angular accelerations of the two bars are equal?
Answer: B
The force must be \(\frac{mg}{2}\).
Both bars are horizontal with F acting downward at the free end, so gravity adds to F in the end-pinned case but has no moment about the centre pin in the second case.
\[ \begin{aligned} \alpha_1 &= \frac{2FL + mgL}{\tfrac{4}{3}mL^{2}} = \frac{3(2F + mg)}{4mL} \\ \alpha_2 &= \frac{FL}{\tfrac{1}{3}mL^{2}} = \frac{3F}{mL} \end{aligned} \]
using \(I = m(2L)^{2}/3\) about an end and \(I = m(2L)^{2}/12\) about the centre. Setting \(\alpha_1 = \alpha_2\) gives \(2F + mg = 4F\), so \(F = mg/2\).
Ignoring gravity would make the two accelerations unequal for any F, and mg, mg/4 and mg/6 do not satisfy the balance above.
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