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Two uniform rods A and B, each of mass m and length l, are connected as follows: rod A is pivoted at its upper end O and hangs vertically. The midpoint of rod B is pinned to the lower end of rod A so that rod B can rotate freely relative to rod A in the vertical plane. For small angular oscillations of the system, the angular frequency is given by sqrt(N*g / (8*l)). Find the integer value of N.

1. 3
2. 6
3. 9
4. 12

Correct answer: 9

Solution

For small oscillations the pin at the midpoint of B forces B's midpoint to follow the tip of A. Since B is free to rotate at the pin, in the equilibrium state B hangs vertically. For small angular displacement theta of rod A: B's midpoint displaces by l*theta, but B itself stays approximately vertical (the pin is frictionless, so no net torque acts on B about its own midpoint), meaning B translates with its midpoint. Thus B's CM moves with the tip of A and is at distance l from O. Restoring torque about O: tau_A = mg*(l/2)*theta, tau_B = mg*l*theta, total tau = (3mgl/2)*theta. Moment of inertia about O: I_A = ml²/3 (rod about end), I_B = m*l² (mass m at distance l, since B translates rigidly). Total I = ml²/3 + ml² = 4ml²/3. omega² = tau/I = (3mgl/2) / (4ml²/3) = (3g/2)*(3/(4l)) = 9g/(8l). So N = 9.

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