Question

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.

Answer 1

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 $I_{oi}$

$I_{oi}$ = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m²  = 0.81 kg.m²

moment of inertia of each object finally $I_{of}$

$I_{of}$  = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²

Now

moment of inertia of student plus stool  $I_{}$ = 5 kg.m²

initial angular speed ω₀ = 0.76 rad/sec

final angular speed ω = ?

Now using conservation of angular momentum;

($I_{}$ + 2 $I_{oi}$ )ω₀ = ($I_{}$ + 2 $I_{of}$ )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) ($I_{}$ + 2 $I_{oi}$ )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) ($I_{}$ + 2 $I_{of}$ )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