Question

A student on a piano stool rotates freely with an angular speed of 3.05 rev/s . The student holds a 1.05 kg mass in each outstretched arm, 0.779 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kg⋅m2 , a value that remains constant.

Calculate the initial kinetic energy of the system.

Calculate the final kinetic energy of the system.

Please list every steps

Answer 1

Given that :

initial angular velocity, _i = 3.05 rev/s

rev/sec change into rad/s -

_i = (3.05) (2)

_i = 19.1 rad/s

mass of object which holds in arm, m₁ = 1.05 kg

distance from the axis of rotation, r₁ = 0.779 m

combined moment of inertia of the student and the stool, I_total = 5.53 kg.m²

The initial kinetic energy of the system which is given as :

K.E_initial = (1/2) I_oi²                                                                { eq.1 }

where, I_o = initial moment of inertia = I_t + m r₁²

inserting the values in above eq.

K.E_initial = (0.5) [(5.53 kg.m²) + 2(1.05 kg) (0.779 m)²] (19.1 rad/s)²

K.E_initial = (0.5) [(5.53 kg.m²) + (1.27 kg.m²)] (19.1 rad/s)²

K.E_initial = (0.5) (6.8 kg.m²) (364.8 rad²/s²)

K.E_initial = 1240.3 J