Answers:
a) [tex]1.05 rad/s[/tex]
b) [tex]1.38 J[/tex]
Explanation:
a) Final angular velocity
:
Before solving this part, we have to stay clear that the angular momentum [tex]L[/tex] is conserved, since we are dealing with circular motion, then:
[tex]L_{o}=L_{f}[/tex]
Hence:
[tex]I_{i} \omega_{i}=I_{f} \omega_{f}[/tex] (1)
Where:
[tex]I_{i}[/tex] is the initial moment of inertia of the system
[tex]\omega_{i}=0.73 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_{i}[/tex] and [tex]I_{f}[/tex]:
[tex]I_{i}=I_{s}+2mr_{i}^{2}[/tex] (2)
[tex]I_{f}=I_{s}+2mr_{f}^{2}[/tex] (3)
Where:
[tex]I_{s}=8 kgm^{2}[/tex] is the student's moment of inertia
[tex]m=2 kg[/tex] is the mass of each object
[tex]r_{i}=1 m[/tex] is the initial radius
[tex]r_{f}=0.28 m[/tex] is the final radius
Then:
[tex]I_{i}=8 kgm^{2}+2(2 kg)(1 m)^{2}=12 kgm^{2}[/tex] (4)
[tex]I_{f}=8 kgm^{2}+2(2 kg)(0.28 m)^{2}=8.31 kgm^{2}[/tex] (5)
Substituting the results of (4) and (5) in (1):
[tex](12 kgm^{2}) (0.73 rad/s)=8.31 kgm^{2}\omega_{f}[/tex] (6)
Finding [tex]\omega_{f}[/tex]:
[tex]\omega_{f}=1.05 rad/s[/tex] (7) This is the final angular speed
b) Change in kinetic energy:
The rotational kinetic energy is defined as:
[tex]K=\frac{1}{2}I \omega^{2}[/tex] (8)
And the change in kinetic energy is:
[tex]\Delta K=\frac{1}{2}I_{f} \omega_{f}^{2}-\frac{1}{2}I_{i} \omega_{i}^{2}[/tex] (9)
Since we already calculated these values, we can solve (9):
[tex]\Delta K=\frac{1}{2}(8.31 kgm^{2}) (1.05 rad/s)^{2}-\frac{1}{2}(12 kgm^{2}) (0.73 rad/s)^{2}[/tex] (10)
Finally:
[tex]\Delta K=1.38 J[/tex] This is the change in kinetic energy
