A thin uniform square plate ABCD of side a and mass m is suspended in a vertical plane as shown in the figure. AE and BF are two massless inextensible strings. The line AB is horizontal. The tension in AE just after BF is cut will be
mg
2mg5
2mg7
3mg5
A
3mg5
B
2mg5
C
mg
D
2mg7
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Solution
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net a0 be the acceleration of centre G and α be the angular acceleration of the plate just after cutting the string BF
mg−T=ma0...........(i)
τG=IGα⇒T(92)=ma2σα
⇒T=maα3...........(ii)
Refer above image.
Just after cutting BF, net
acceleration of A is resultant of two acceleration a0 and αa√2 as shown
But the net acc. of A in the vertical direction should be zero.
Hence,
a0=α9√2sin(450)
⇒a0=α92..........(iii)
from (i),(ii) and (iii)
T=2mg5
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