Answer:
a) the new angular speed of the student is 0.9642 rad/s
b)
the kinetic energy of the student before the objects are pulled in is 1.9119 J
the kinetic energy of the student after the objects are pulled in is 2.4252 J
Explanation:
Given that;
mass of each object m = 1 kg
distance of objects from axis of rotation r = 0.9 m
Moment of inertia of each object initially [tex]I_{oi}[/tex]
[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m² = 0.81 kg.m²
moment of inertia of each object finally [tex]I_{of}[/tex]
[tex]I_{of}[/tex] = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²
Now
moment of inertia of student plus stool [tex]I_{}[/tex] = 5 kg.m²
initial angular speed ω₀ = 0.76 rad/sec
final angular speed ω = ?
Now using conservation of angular momentum;
([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω
so we substitute
(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω
5.0312 = 5.2178 ω
ω = 5.0312 / 5.2178
ω = 0.9642 rad/s
Therefore, the new angular speed of the student is 0.9642 rad/s
b)
K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²
= (0.5) (5 + 2 (0.81) )(0.76)²
= 0.5 × 6.62 × 0.5776
= 1.9119 J
Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J
KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²
= (0.5) (5 + 2 (0.1089) ) (0.9642)²
= 0.5 × 5.2178 × 0.9296
= 2.4252 J
Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J
