A thin uniform rod of mass m and length L is hinged at its upper end and released from rest in horizontal position. The tension at a point located at a distance L/3 from the hinge point , when the rod becomes vertical, will be -
A
22mg27
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B
11mg13
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C
6mg11
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D
2mg
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Solution
The correct option is D2mg When the rod becomes vertical, apply conservation of energy.
mgL2=12mL23ω2 ω=√3g/L Centre of mass of the lower two-third part moves in a circle of radius 2L/3. We apply Newton’s second law on this part T−2mg3=(2m3)(2L3)ω2(or)T=2mg