A ballet dancer is rotating about his own vertical axis on smooth horizontal floor with a time period 0.5 sec. 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

Question

A ballet dancer is rotating about his own vertical axis on smooth horizontal floor with a time period 0.5 sec. 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) 0.1 sec (B) 0.25 sec (C) 0.32 sec (D) 0.4 sec

Tardigrade Admin · Accepted Answer

I1ω 1=I2ω 2 MK12(2 π /T1)=MK22(2 π /T2) (T2/T1) = (. (K2/K1) .)2 T2 = 0.64 Ã 0.5 = 0.32 sec

Q. A ballet dancer is rotating about his own vertical axis on smooth horizontal floor with a time period 0.5 sec. 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

0.1 sec

0.25 sec

0.32 sec

0.4 sec

$I_{1} ω_{1} = I_{2} ω_{2}$
$M K_{1}^{2} \frac{2 π}{T _{1}} = M K_{2}^{2} \frac{2 π}{T _{2}}$

$\frac{T _{2}}{T _{1}} = (\frac{K _{2}}{K _{1}})^{2}$

T₂ = 0.64 × 0.5

= 0.32 sec

Q. A ballet dancer is rotating about his own vertical axis on smooth horizontal floor with a time period 0.5 sec. 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

0.1 sec

0.25 sec

0.32 sec

0.4 sec

I1​ω1​=I2​ω2​ MK12​T1​2π​=MK22​T2​2π​ T1​T2​​=(K1​K2​​)2 T2 = 0.64 × 0.5 = 0.32 sec

$I_{1} ω_{1} = I_{2} ω_{2}$
$M K_{1}^{2} \frac{2 π}{T _{1}} = M K_{2}^{2} \frac{2 π}{T _{2}}$

$\frac{T _{2}}{T _{1}} = (\frac{K _{2}}{K _{1}})^{2}$

T₂ = 0.64 × 0.5

= 0.32 sec