Answer :
Answer:
The ratio is 0.53
Explanation:
There aren't external torques on the ice skater because we can assume ice surface frictionless and there're not external forces, that implies angular moment is conserved so it has the same value before and after him pulling his arms. Angular momentum is the product between angular velocity (ω) and moment of inertia (I).
The initial angular moment is:
[tex]L=I_{i}\omega_{i} [/tex] (1)
and final angular moment is:
[tex] L=I_{f}\omega_{f} [/tex] (2)
Because conservation of angular moment we can equate (1) and (2):
[tex]I_{i}\omega_{i}=I_{f}\omega_{f} [/tex]
rearranging the expression:
[tex]\frac{I_{f}}{I_{i}}=\frac{\omega_{i}}{\omega_{f}} [/tex]
So, the ratio of the skater's final moment of inertia to his initial moment of inertia is:
[tex] \frac{I_{f}}{I_{i}}=\frac{3.17\frac{rad}{s}}{5.96\frac{rad}{s}}[/tex]
[tex] \frac{I_{f}}{I_{i}}=0.53[/tex]
The required ratio of of the skater's final moment of inertia to his initial moment of inertia is 0.53 : 1.
Given data:
The initial angular speed of the ice skater is, [tex]\omega = 3.17 \;\rm rad/s[/tex].
The final angular speed of the ice skater is, [tex]\omega ' =5.96 \;\rm rad/s[/tex].
In this problem, we can apply the conservation of the angular momentum, which says that in the absence of external torque, the initial angular momentum is equal to the final angular momentum.
Initial angular momentum = final angular momentum
[tex]L = L'\\I \times \omega = I' \times \omega'[/tex]
Here,
I is the initial moment of inertia.
I' is the final moment of inertia.
Solving as,
[tex]\dfrac{I'}{I} = \dfrac{ \omega}{ \omega'} \\\\\dfrac{I'}{I} = \dfrac{3.17}{ 5.96}\\\\\dfrac{I'}{I} = 0.53[/tex]
Thus, the required ratio of of the skater's final moment of inertia to his initial moment of inertia is 0.53 : 1.
Learn more about the angular momentum here:
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