The angular momentum of the skater is given by

`L =Iomega`

where `omega` is the angular velocity in m/s and `I` is the moment of inertia.

If a body is made up of point masses `m_i` `i=1,2,3,...,n` at radius

`r_i` from the axis of rotation, the moment of inertia is...

## See

This Answer NowStart your **subscription** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The angular momentum of the skater is given by

`L =Iomega`

where `omega` is the angular velocity in m/s and `I` is the moment of inertia.

If a body is made up of point masses `m_i` `i=1,2,3,...,n` at radius

`r_i` from the axis of rotation, the moment of inertia is given by

`I = sum_1^n m_ir_i^2`

Therefore, for larger bodies, the moment of inertia is larger relative to mass.

When the skater's arms are outstretched her moment of inertia is larger, therefore a larger angular momentum is required to keep her going at the same angular velocity. As momentum is conserved, the skater will spin faster when she brings her arms in since the angular momentum will stay the same, but the moment of inertia will decrease.

**She spins faster when her arms are brought in because her angular momentum remains the same, but her moment of inertia decreas****es**