Calculate a diver's turns in air

[agadag]

Homework Statement

The rotational inertia for a diver in a pike position is about 15.5 kg m2; it is only 8.0 kg m2 in a tucked position (the figure above).

(a) If the diver gives himself an initial angular momentum of 106 kg m2/s as he jumps off the board, how many turns can he make when jumping off a 10.0-m platform in a tuck position? [Hint: Gravity exerts no torque on the person as he falls; assume he is rotating throughout the 10.0 m dive.]

3.0
4.0
2.0
5.0
4.5
(b) How many in a pike position?

1.2
1.8
1.6
1.1
1.4

Homework Equations

W_iI_i= W_fI_f
n= (w_f/2Pie) t

The Attempt at a Solution

The teacher mildly went over this problem in class. These are the eqns he gave us. I tried to solve for w_f and substituted that answer in the second eqn to find the number of turns in time t, which i calculated thru the eqn :
h=v_it + 1/2 at²
I really have NO idea how to do this problem, i only know these eqns from wat the instructor provided. If you can please teach me how to go about solving the problem and understanding it, i would really truly appreciate it! Thanks!

 

[Delphi51]

Don't you just need angular momentum = I*ω to find ω?
After that, use your second formula to get the number of turns in the time it takes to fall to the water.

 

[kerimek]

First, let's define some variables:
Ii = initial moment of inertia (given as 8.0 kg m^2 for tucked position)
If = final moment of inertia (given as 15.5 kg m^2 for pike position)
Wi = initial angular momentum (given as 106 kg m^2/s)
Wf = final angular momentum (unknown)
t = time (unknown)
n = number of turns (unknown)

To solve for Wf, we can use the equation WiIi = WfIf and rearrange it to solve for Wf:
Wf = (WiIi)/If
Substituting in the given values, we get:
Wf = (106 kg m^2/s * 8.0 kg m^2) / 15.5 kg m^2
Wf = 54.19 kg m^2/s

Next, we can use the equation n = (Wf/2πIi) * t to solve for the number of turns, where t is the time it takes for the diver to complete the 10.0 m dive. We can calculate t using the equation h = vit + 1/2 at^2, where h is the height of the platform (10.0 m), vi is the initial velocity (0 m/s since the diver is starting from rest), and a is the acceleration due to gravity (-9.8 m/s^2). Solving for t, we get:
t = √(2h/a)
t = √(2*10.0 m / -9.8 m/s^2)
t = 1.43 s

Now, we can plug in the values for Wf, Ii, and t to solve for n:
n = (54.19 kg m^2/s / 2π * 8.0 kg m^2) * 1.43 s
n = 1.81 turns

Therefore, the diver can make approximately 1.8 turns in a pike position when jumping off a 10.0 m platform with an initial angular momentum of 106 kg m^2/s. Similarly, you can use the same method to solve for the number of turns in a tucked position.