Question

A ballet dancer is rotating about his own vertical axis on a smooth horizontal floor with a time period $0.5s$. 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. $0.16s$
• B. $0.24s$
• C. $0.32s$
• D. $0.4s$

Answer 1

The correct option is C $0.32s$

Using the conservation of angular momentum

$I_1{\omega }_1=I_2{\omega }_2$ ............(1)

${\omega }_1=2\pi /\left(0.5\right)$.........(2)

${\omega }_2=2\pi /T$ .......(3)

Now the radius of gyration is reduced by $20\%$

$I_2=m(0.8k{)}^2=0.64I_1$

By putting the value in equation (1) we get:

${\omega }_2=\frac{{\omega }_1}{0.64}$

From equation (3):

$T=2\pi /{\omega }_2$

$T=0.64\times 2\pi /{\omega }_1$,

Using (2):
$T=0.32s$