A diver in midair has an angular velocity of 6.0 rad/s and a moment of inertia of 1.2 kg·m2. He then pulls is arms and legs into a tuck position and his angular velocity increases to 12 rad/s. The net external torque acting on the diver is zero. What is his moment of inertia in the tuck position?

The Conservation of Angular Momentum will be conserved. (Hint: it states that the net external torque acting on the diver is zero.)

Angular Momentum =
moment of inertia x angular velocity

(6.0) (1.2) = 12x
7.2 = 12x
x = 0.6 kg·m^2

Ur dumb :3

0.6

Well, well, well, look at our acrobatic friend here! Quite the daredevil, huh?

Now, let's get down to business. We know that angular momentum is conserved when the net external torque acting on an object is zero. So, in both the original and tuck positions, the angular momentum remains the same.

Angular momentum (L) is given by the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. In the original position, L = 1.2 kg·m² * 6.0 rad/s.

Since angular momentum is conserved, in the tuck position, L = I' * 12 rad/s, where I' is the moment of inertia in the tuck position.

Setting up the equation:

1.2 kg·m² * 6.0 rad/s = I' * 12 rad/s

Now, we can solve for I' by rearranging the equation:

I' = (1.2 kg·m² * 6.0 rad/s) / 12 rad/s

Doing some quick math, we find that I' = 0.6 kg·m².

So, in the tuck position, our jolly diver's moment of inertia is 0.6 kg·m². Happy tucking, my friend! But remember, no clowning around while doing dangerous stunts!

To find the moment of inertia of the diver in the tuck position, we can use the law of conservation of angular momentum, which states that the angular momentum of a system remains constant if no external torques act on it.

The angular momentum of an object is given by the product of its moment of inertia (I) and its angular velocity (ω). Mathematically, it can be represented as:

L = I * ω

Since the net external torque acting on the diver is zero, the angular momentum should remain constant as he changes his body position.

Initially, the angular velocity of the diver is 6.0 rad/s, and his moment of inertia is 1.2 kg·m^2. Let's call this moment of inertia I1.

L1 = I1 * ω1

When the diver pulls his arms and legs into a tuck position, his angular velocity increases to 12 rad/s. Let's call the moment of inertia in the tuck position I2.

L2 = I2 * ω2

Since the angular momentum remains constant, we can equate L1 and L2:

L1 = L2

I1 * ω1 = I2 * ω2

Substituting the given values:

1.2 kg·m^2 * 6.0 rad/s = I2 * 12 rad/s

Simplifying the equation, we find:

7.2 = 12 I2

Dividing both sides by 12:

I2 = 7.2 / 12

Therefore, the moment of inertia of the diver in the tuck position is 0.6 kg·m^2.