Question

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 5.2 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is $1.56\frac{rev}{s}$

Explanation:

Initial moment of inertia when arms and legs in is $I_i=0.90 kg.m^{2}$

Final moment of inertia when her arms and on leg open outward, $I_f=3.0 kg.m^{2}$

Initial angular speed $w_i=5.2\frac{rev}{s}$

Let the final angular speed be $w_f$

Since external torque on her is zero so we can apply conservation of angular momentum

$\therefore L_f=L_i$

=>$I_fw_f=I_iw_i$

=>$w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}$

Thus her angular speed (in rev/s) when her arms and one leg open outward is $1.56\frac{rev}{s}$