Question

A uniform metre scale of length 1 m is balanced on a fixed semi-circular cylinder of radius 30 cm as shown in figure. One end of the scale is slightly depressed and released. The time period (in seconds) of the resulting simple harmonic motion is (Take g=10 ms−2) ππ2π3π4

Solution

The correct option is C π3 Let the scale be displaced by a small angle θ. From the figure, the magnitude of the restoring torque about A = force × perpendicular distance. τrest=mg×AB=mg×Rsinθ Since θ is small, sinθ≃θ. ⇒Id2θdt2=−mg×Rθ ⇒d2θdt2=(−mgRI)θ Comparing with SHM equation α=−ω2θ, we get ω=√mgRI  ⇒2πT=√mgRI  ⇒T=2π√ImgR ⇒T=πL√3gR     [∵I=mL212] Using the values L=1 m, g=10 ms−2 and R=0.3 m, we get T=π3 sec. Hence, option (c) is the correct answer.

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