A small ball of mass $M$ is attached to the end of a uniform rod of equal mass $M$ and length $L$ that is pivoted at the top. Determine the tensions in the rod (a) at the pivot and (b) at the point $P$ when the system is stationary. (c) Calculate the period of oscillation for small displacements from equilibrium and (d) determine this period for $L$ = $2.00 m$ .

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Solution

(a) At the pivot , $F = M g + M g = 2 M g$
(b) $F = M g (\frac{y}{L}) + M g = M g (1 + \frac{y}{L})$
(c) $T = \frac{4 π}{3} \sqrt{\frac{2 L}{g}}$
(d) For $L = 2.00 m, T = 2.68 s$

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(a) At the pivot , F=Mg+Mg=2 Mg(b) F=Mg(yL)+Mg=Mg(1+yL)(c) T=4π3√2Lg(d) For L=2.00 m,T=2.68 s

(a) At the pivot , $F = M g + M g = 2 M g$
(b) $F = M g (\frac{y}{L}) + M g = M g (1 + \frac{y}{L})$
(c) $T = \frac{4 π}{3} \sqrt{\frac{2 L}{g}}$
(d) For $L = 2.00 m, T = 2.68 s$