Question

When a diver gets into a tuck position by pulling in her arms and legs, she increases her angular speed. Before she goes into the tuck position, her angular velocity is 5.5 rad/s, and she has a moment of inertia of 1.4 kg m2. Once she gets into the tuck position, her angular speed is 14.5 rad/s. Determine her moment of inertia, in kg m2, when she is in the tuck position. Assume the net torque on her is zero?

Answer 1

The moment of inertia of the diver when she is in the tuck position is $0.53 kg m^2$

Explanation:

When the net torque acting on an object is zero, the angular momentum of the object must be conserved.

The angular momentum is given by:

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular velocity

Therefore, in this situation we can write:

$L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2$

where:

$I_1 = 1.4 kg m^2$ is the initial moment of inertia of the diver

$\omega_1 = 5.5 rad/s$ is the initial angular velocity

$I_2$ is the final angular momentum of the diver

$\omega_2 = 14.5 rad/s$ is the final angular velocity of the diver

Solving the equation for $I_2$, we find:

$I_2 = \frac{I_1 \omega_1}{\omega_2}=\frac{(1.4)(5.5)}{14.5}=0.53 kg m^2$

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