A thin uniform rod of mass M and length L is hinged at an end and released from rest in the horizontal position. The tension at a point located at a distance L/3 from the hinged point, when the rod becomes vertical, will be
22Mg/27
11Mg/13
6Mg/11
2Mg
A
2Mg
B
11Mg/13
C
22Mg/27
D
6Mg/11
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Solution
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2mg
0+mg×0=12Iw2+mg(−L2)
12(mL23)w2=mgL2
w=√3gL
T=mg+mv2r
=m[g+rw2]
=m[g+2Lg×3g2]
=2m3(g+2g)
=2mg
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