A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to `7.6 kg m^(2)`.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Explanation

(a) The moment of inertia of a system of man and rotating platform is `I=7.6kgm^(2)`
The moment of inertia of system when the man stretches his hands to a distance of 90 cm
`I.=2mr^(2)`
`=2xx5xx(0.9)^(2)` where `m=5kg`
`=8.1 kgm^(2) r=90cm=0.9m`
`therefore` Initial moment of inertia of the system
`I_(i)=I+I.`
`=7.6+8.1`
`=15.7kgm^(2)`
and initial angular speed `omega_(i)=30` rpm
`therefore` initial angular momentum of system
`L_(i)=I_(i)omega_(i)`
`=15.7xx30`
`=471 Js...(1)`
moment of inertia when the man holds his hands to a distance of 20 cm
`therefore I.=2xxm(r.)^(2)`
`=2xx5xx(0.2)^(2)[because r.=20cm=0.2m]`
`=0.4kgm^(2)`
`therefore` final moment of inertia of system
`I_(f)=I+I.`
`therefore I_(f)=7.6+0.4`
`=8kgm^(2)`
Suppose final angular velocity is `omega_(f)`
`therefore` Final angular momentum
`L_(f)=I_(f)omega_(f)`
`therefore L_(f)=8omega_(f)" "...(2)`
Since no external torque is acting
`L_(i)=L_(f)`
`471-8omega_(f)`
`therefore omega_(f)=(471)/(8)`
`=58.88rpm~~59rpm`
(b) No, kinetic energy is not conserved in the given process
`omega_(i)ltomega_(f)`
`K_(i)ltK_(f)[because K=(1)/(2)Iomega^(2)]`
Kinetic energy increases. This additional kinetic energy comes from the work done by the man to fold his hands towards himself.