Question

A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves thediving board in the straight position (with some unknown angularvelocity), he pulls himself into a ball as closely as possible (thetuck position) and makes 2.00 complete rotations in 1.33 s. If hisrotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angularvelocity when he left the diving board?

Answer 1

3.14946 rad/s

Step-by-step explanation:

`I_i` = Intial moment of inertia

`I_f` = Final moment of inertia

`\omega_i` = Initial angular velocity

`\omega_f` = Final angular velocity =
`(2)/(1.33)* 2\pi\ rad/s`

`(I_f)/(I_i)=(1)/(3)`

In this system the angular momentum is conserved

`L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=(I_f\omega_f)/(I_i)\\\Rightarrow \omega_i=(1* (2)/(1.33)* 2\pi)/(3)\\\Rightarrow \omega_i=3.14946\ rad/s`

The angular velocity when the diver left the board is 3.14946 rad/s