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Question

A ballet dancer is rotating about his own vertical axis on a smooth horizontal floor with a time period 0.5 s. The dancer folds himself close to his axis of rotation due to which his radius of gyration decreases by 20%, then his time period is :

A
0.16 s
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B
0.24 s
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C
0.32 s
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D
0.4 s
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Solution

The correct option is C 0.32 s
Using the conservation of angular momentum
I1ω1=I2ω2 ............(1)
ω1=2π/(0.5).........(2)
ω2=2π/T .......(3)
Now the radius of gyration is reduced by 20%
I2=m(0.8k)2=0.64I1
By putting the value in equation (1) we get:
ω2=ω10.64
From equation (3):
T=2π/ω2
T=0.64×2π/ω1,
Using (2):
T=0.32 s

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