A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so th

Question

denis-greek [22]

4 years ago

13

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so th

e moment of inertia decreases to 1.80 kg·m2

Physics

1 answer:

harkovskaia [24] · Answer 1 · 2021-12-28T10:42:06+03:00

Answer:

The angular velocity when the arms are pulled is ω₂= 6.25 rad/s

Explanation:

Assuming that you want to determine the final angular velocity when is the arms are pulled, then it is calculated using the principle of conservation of angular momentum. It states that:

I₁ω₁=I₂ω₂

where I = moment of inertia , ω= angular velocity , 1 and 2 denote the skater with extended hands and pulled respectively.

Thus

I₁ω₁=I₂ω₂

ω₂= I₁ω₁/I₂

replacing values

ω₂= ω₁ *(I₁/I₂) = 5 rad/s *(2.25 kg·m2/1.80 kg·m2) = 6.25 rad/s

ω₂= 6.25 rad/s