Question

A student sits on a rotating stool holding two 2.8-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is

3.0 kg · m²

and is assumed to be constant. The student then pulls in the objects horizontally to 0.50 m from the rotation axis.

(a) Find the new angular speed of the student.
rad/s

(b) Find the kinetic energy of the student before and after the objects are pulled in.

before	J
after	J

Answer 1

As given moment of inertia mi=3.0kgm²

Mass M=2.8kg, initially length l=1.0m, l₂=0.50m,

angular speed at begining w_i=0.75rad/s, w_f=?

Now total moment of inertia at beginning is mi=3+2×(2.8×1.0²)=8.6kgm²

Moment of inertia final

Mi=3+2(2.8×0.50²)=4.4kgm²

Angular momentum at beginning

L= moment of inertia at beginning ×angular speed at beginning

L_i=8.6×0.75=6.45kgm²/s

Angular momentum at final

L_f=mi_final×w_f

L_f=4.4×w_f

angular momentum of body is conserve so

L_i=Lf

6.45=4.4×w_f

W_f=6.45/4.4=1.46rad/s

therefore nenew angular speed=1.46rad/s

B) now kinetic energy KE_i=1/2×mi×w_i²

KE_i=0.5×8.6×0.75²

KE_i=2.41J

Kinetic energy final

KE_f=0.5×4.4×1.46²=4.68J