Question

A man stands at the centre of a circular platform holding his arms extended horizontally with 4 kg block in each hand. He is set rotating about a vertical axis at `0.5rev//s`. The moment of inertia of the man plus platform is `1.6kg-m^(2)`, assumed constant, the block are 90 cm from the axis of rotation. He now pulls the blocks in towards his body until they are 15 cm from the axis of rotation. Find (a) his new angular velocity and (b) the initial and final kinetic energy of the man and platform (c) how much work most the man do to pull int he blocks?

Answer 1

(a). `I_(1)omega_(1)=I_(2)omega_(2)`
`thereforeomega_(2)=(I_(1))/(I_(2))omega_(1)`
`=([1.6+(2)(4)(0.9)^(2)])/([1.6+(2)(4)(0.15)^(2)])xx0.5`
`=((8.08)/(1.78))(0.5)`
`=2.27rev//s`
`=14.3rad//s`
(b) `K_(i)=(1)/(2)I_(1)omega_(1)^(2)`
`=(1)/(2)(1.6+(2)(4)(0.9)^(2)0.5xx2pi)^(2)`
`=39.9J`
`K_(f)=(1)/(2)[1.6+(2)(4)(0.15)^(2)](14.3)^(2)`
`=181J`
(c). `W=K_(f)-K_(i)=141.1J`