A trapeze artist performs an aerial maneuver While in a tuck

A trapeze artist performs an aerial maneuver. While in a tucked position, as shown in Figure A, she rotates about her center of mass at a rate of 6.03 rad/s. Her moment of inertia about this axis is 18.5 kg- m^2. A short time later the aerialist is in the straight position, as shown in Figure B. If the moment of inertia about her center of mass in this position is now 31.3 kg- m^2, what is her rotational speed?

Solution

Answer :

Given that,

I₁ = Moment of Inertia (in tucked position) = 18.5 Kg.m²

₁ = angular velocity (in tucked position) = 6.03 rad/s

I₂ = Moment of Inertia (in straight position) = 31.3 Kg.m²

₂ = angular velocity (in straight position) = to be determined

We know that Angular momentum L = I.

In this case, angular momentum is conserved since no external force is acting on the Trapeze artist.

Therefore, L₁ = L₂ . (Before and after)

which implies that,    I₁₁ = I₂₂

₂ = I₁₁/ I₂ = (18.5 Kg.m²) (6.03 rad/s) / (31.3 Kg.m²) = 3.564 rad/s.

This is the required answer.