A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.16 complete rotations in 1.31 s. If his rotational inertia decreases by a factor of 2.90 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?

Question

A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.16 complete rotations in 1.31 s. If his rotational inertia decreases by a factor of 2.90 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?

Answer 1

The thing that stays constant here is the angular momentum which is angular velocity (omega) times rotational inertia (I)

so write:

angular velocity before times rotational inertia before equals angular velocity after times rotational inertia after.
By the way angular velocity is usually expressed in radians per second so you will have to get that from the "after" data of 2.16 rotations in 1.31 seconds.

Answer 2

To solve this problem, we can use the principle of conservation of angular momentum. The angular momentum of an object is conserved when there is no external torque acting on it. In this case, we can assume that there is no external torque acting on the diver during the rotations.

The formula for angular momentum is given by:

L = Iω

Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

We are given that the diver makes 2.16 complete rotations in 1.31 seconds. A complete rotation is equal to 2π radians. Therefore, the angle rotated by the diver can be calculated as:

θ = 2.16 * 2π

Next, we are told that the rotational inertia decreases by a factor of 2.90 when the diver goes from the straight position to the tuck position. This means that the moment of inertia in the tuck position (Ituck) is equal to 1/2.90 times the moment of inertia in the straight position (Istraight).

So, Ituck = (1/2.90) * Istraight

Now, let's substitute the formulas and values into the equation for angular momentum:

Lstraight = Istraight * ωstraight
Ltuck = Ituck * ωtuck

Since angular momentum is conserved, we can equate the two expressions:

Lstraight = Ltuck

Istraight * ωstraight = Ituck * ωtuck

Substituting the equation for Ituck, we have:

Istraight * ωstraight = (1/2.90) * Istraight * ωtuck

Istraight cancels out:

ωstraight = (1/2.90) * ωtuck

Finally, we can substitute the given values to solve for ωstraight:

ωstraight = (1/2.90) * 2π * 2.16 / 1.31

Calculating this expression, we can find the value of ωstraight, which is the angular velocity when the diver left the diving board.