A trapeze artist sailing through the air rotates with a rotational speed of 0.9·rev/s.

If she contracts to reduce her rotational inertia to one sixth of what it was, what will her new rotational speed be? _____ rev/s?

Can she completely stop her rotation while still airborne? Choose one answer only.
1. yes, she must increase her rotational inertia
2. yes, she must decrease her rotational inertia
3. no, she cannot stop rotating
4. any of these are possible, depending on the skill of the artist

yes

To find the new rotational speed, we can apply the conservation of angular momentum. The formula for angular momentum is given by:

L = Iω

where L represents the angular momentum, I represents the rotational inertia, and ω represents the rotational speed.

Since the trapeze artist contracts to reduce her rotational inertia to one-sixth of what it was, we can write the equation as follows:

L1 = (1/6)I0ω1

Where L1 represents the new angular momentum, I0 represents the initial rotational inertia, and ω1 represents the new rotational speed.

However, the angular momentum (L) is conserved, which means it remains the same before and after the contraction:

L1 = L0

Therefore, we can equate the two equations:

(1/6)I0ω1 = I0ω0

Dividing both sides of the equation by I0:

(1/6)ω1 = ω0

Simplifying the equation, we find:

ω1 = 6ω0

So the new rotational speed (ω1) will be six times the initial rotational speed (ω0).

To answer the second part of the question, can she completely stop her rotation while still airborne, the correct answer is: 3. no, she cannot stop rotating.

Because of the conservation of angular momentum, an object cannot change its rotational speed without an external torque acting upon it. In this case, the trapeze artist does not have any means to exert an external torque on herself while in mid-air, so she cannot stop rotating.