A rod OA of uniform linear mass density and length ‘L’ is hinged at O ...
Problem Statement
A rod OA of uniform linear mass density and length ‘L’ is hinged at O in a vertical plane. Another square plate of same mass is attached with a point A of the rod as shown in the figure.
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Case (i): Square plate is free to rotate about point A
In this case, the square plate attached to point A of the rod can rotate freely about the point. When the system is allowed to oscillate in the vertical plane such that the axis of rotation is horizontal and passing through point O, the time period for the oscillation can be calculated as:
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Case (ii): Square plate is fixed at point A
In this case, the square plate attached to point A of the rod is fixed to the point. When the system is allowed to oscillate in the vertical plane such that the axis of rotation is horizontal and passing through point O, the time period for the oscillation can be calculated as:
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Explanation
The time period for the oscillation in case (i) is greater than the time period for oscillation in case (ii). This is because in case (i), the square plate is allowed to freely rotate about point A, which increases the moment of inertia of the system and hence increases the time period of oscillation. In contrast, in case (ii), the square plate is fixed to point A, which decreases the moment of inertia of the system and hence decreases the time period of oscillation.
Conclusion
In summary, the time period for the oscillation of a system consisting of a rod OA hinged at O in a vertical plane with a square plate attached to point A of the rod depends on whether the square plate is free to rotate about point A or fixed to the point. The time period for oscillation is greater when the square plate is free to rotate and smaller when it is fixed to the point.