Question

Diana, a figure skater, is initially spinning at an angular speed 2.50 rev/s, with her arms and legs inward. Assume that she is a uniform cylinder with a height of 1.4 m, a radius of 18 cm, and a mass of 55 kg. Assume no external torques act.

a) What is her moment of inertia?

b) If she extends her arms outward, what is her new moment of inertia? Assume that

her armspan is 1.3 m and her arms are 8 kg.

c) What is Diana’s new angular speed?

d) Was kinetic energy conserved? Explain.

Answer 1

part a:

moment of inertia=0.5*mass*radius^2=0.5*55*0.18^2=0.891 kg.m^2

part b:

her arms can be modelled as rods of length 0.5*(1.3-2*0.18)=0.5*0.94 m=0.47 m

then moment of inertia of each arm =(1/12)*mass*length^2+mass *distance from the axis^2

distance from axis=half of arm length+radius of the cylinder=(0.47/2)+0.18=0.415 m

then moment of inertia of each arm=(1/12)*8*0.47^2+8*0.415^2=1.5251 kg.m^2

so total moment of inerrtia=0.891+2*1.5251=3.9412 kg.m^2

part c:

as total angular momentum will be conserved,

new angular speed=old angular speed*old moment of inertia/new moment of inertia

=2.5*0891/3.9412=0.5651 rev/s

part d:

initial kinetic energy=0.5*moment of inertia*angular speed^2

=0.5*0.891*2.5^2=2.784

final kinetic energy=0.5*3.9412*0.5651^2=0.62928

hence energy is not conserved.

energy is lost in work done in moving the hands up