Answers:
a) [tex]1.17 rad/s[/tex]
b) [tex]1.31 J[/tex]
Explanation:
a) We can solve this knowing the angular momentum [tex]L[/tex] is conserved, then:
[tex]L_{o}=L_{f}[/tex]
[tex]I_{o} \omega_{o}=I_{f} \omega_{f}[/tex] (1)
Where:
[tex]I_{o}[/tex] is the initial moment of inertia of the system
[tex]\omega_{o}=0.69 rad/s[/tex] is the initial angular velocity
[tex]I_{f}[/tex] is the final moment of inertia of the system
[tex]\omega_{f}[/tex] is the final angular velocity
But first, we have to find [tex]I_{o}[/tex] and [tex]I_{f}[/tex]:
[tex]I_{o}=I_{s}+2mr_{o}^{2}[/tex] (2)
[tex]I_{f}=I_{s}+2mr_{f}^{2}[/tex] (3)
Where:
[tex]I_{s}=4 kgm^{2}[/tex] is the student's moment of inertia
[tex]m=4 kg[/tex] is the mass of each object
[tex]r_{o}=0.7 m[/tex] is the initial radius
[tex]r_{f}=0.29 m[/tex] is the final radius
Then:
[tex]I_{o}=4 kgm^{2}+2(4 kg)(0.7 m)^{2}=7.92 kgm^{2}[/tex] (4)
[tex]I_{f}=4 kgm^{2}+2(4 kg)(0.29 m)^{2}=4.67 kgm^{2}[/tex] (5)
Substituting the results of (4) and (5) in (1):
[tex](7.92 kgm^{2}) (0.69 rad/s)=4.67 kgm^{2}\omega_{f}[/tex] (6)
Finding [tex]\omega_{f}[/tex]:
[tex]\omega_{f}=1.17 rad/s[/tex] (7) This is the final angular velocity
b) The rotational kinetic energy is:
[tex]K=\frac{1}{2}I \omega^{2}[/tex] (8)
The change in kinetic energy is:
[tex]\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{o} \omega_{o}^{2}[/tex] (9)
Since we already calculated these values, we can solve (9):
[tex]\Delta K=\frac{1}{2}(4.67 kgm^{2}) (1.17 rad/s)^{2}-\frac{1}{2}(7.92 kgm^{2}) (0.69 rad/s)^{2}[/tex] (10)
Finally:
[tex]\Delta K=1.31 J[/tex]
