A skater has outstretched arms preparing for a turn with a moment of inertia of 2.56 kg m2. When the arms are joined the body moment of inertia around the axis of
rotation is equal to 0.40 kg m2. If the original angular speed of the skater is 0.40 rev/s, what is the final angular speed?

The final angular speed is 16.1 rad/s

Explanation:

Given;

initial moment of inertia, I₁ = 2.56 kg.m²

final moment of inertia, I₂ = 0.40 kg.m²

initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s

Apply the principle of conservation of angular momentum;

I₁ω₁ = I₂ω₂

where;

ω₂ is the final angular speed

ω₂ = (I₁ω₁) / (I₂)

ω₂ = (2.56 x 2.514) / (0.4)

ω₂ = 16.1 rad/s

Therefore, the final angular speed is 16.1 rad/s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be