A BUG QUESTION
A square metal frame in the vertical plane is hinged at O at its center. A bug of mass = 0.5 gram moves to and fro about M along the rod PN which is at a perpendicular distance of 100 cm from the hinge such that whole frame is always stationary, even though the frame is free to rotate in vertical plane. M is the midpoint of rod PN. (take (3.14)^2 = 10) Find the time period(in seconds) of the motion of the bug.
The answer is 2.
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1 solution
Friction is what causes the frame to stay in equilibrium. The torque produced by the friction= fL (f is the friction and L=100cm=1m) which balances the torque produced by the weight of the bug. Now consider the bug at a distance x from the mean position(which is right below O on the rod the bug moves on). As the system is in equilibrium fL=mgx (no net torque). => f = mgx/L ........... (1) now this friction is the restoring force => f = (mw^2)x ............. (2) equating (1) and (2) w=sqrt{g/L} the time period = 2pi/w => T = 2pisqrt{L/g} => T = 2 seconds (g = 10m/s^2)