Question

A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed

Answer 1

The new angular speed is
`w = 6 \ rev/s`

Step-by-step explanation:

From the question we are told that

The angular velocity of the spin is
`w_o = 3 \ rev/s`

The original moment of inertia is
`I_o`

The new moment of inertia is
`I =(I_o)/(2)`

Generally angular momentum is mathematically represented as

`L = I * w`

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

`I * w = constant`

=>
`I_o * w _o = I * w`

where w is the new angular speed

So

`I_o * 3 = (I_o)/(2) * w`

=>
`w = (3 * I_o)/((I_o)/(2) )`

=>
`w = 6 \ rev/s`