A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec. The moment of inertia of the student plus the stool is 5 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis. Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.

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

= mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m²  = 0.81 kg.m²

moment of inertia of each object finally

 = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²

Now

moment of inertia of student plus stool   = 5 kg.m²

initial angular speed ω₀ = 0.76 rad/sec

final angular speed ω = ?

Now using conservation of angular momentum;

( + 2 )ω₀ = ( + 2 )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) ( + 2 )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) ( + 2 )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

found in the nucleus

has mass of one amu