I think I might be on the right track, if someone could let me know if this is correct I would greatly appreciate it.
A 51 kg diver is spinning forward in a tuck position at 1.9 rev/s, as shown in Fig. 13-38a. In this position, the diver's rotational inertia is
3.3 kg·m2. At the peak of her trajectory, the diver suddenly straightens out in a horizontal position as shown in Fig. 13-38b. In this new
position, the diver has an overall length of 2.2 m with center of mass midway along this length, and a rotational inertia of 21 kg·m2 about
the rotation axis. What is the minimum height of the diver above the water if she is to enter the water hands first? Neglect the diver's horizontal translational motion.
Basically, picture a and b look similar---on the bottom, both have the "diving board" on the left, and the water line across the rest of the bottom, with the diver "jumping" to the right. Both have a dotted trajectory line following the divor, looks sorta parabolic. The difference is in A, the diver is nearly in the curled/ball position, while in B, she's at the "top of the trajectory" tangent to the curve of the trajectory, basically her arms and legs (thus her whole body) are extended horizontally at the peak of the parabolic trajectory.
So I think she has to "twist" 90 degrees along the trajectory, from hands point horizontally to hands pointing straight down as she "curves" downward and to the right. This would imply a quarter of a revolution. We can use the conservation of angular momentum to find wf,
I0*w0 = If*wf. So rearranging and plugging in gives wf = 3.3*1.9 / 21 = 0.2986, which is rev/s. So a full revolution would take 1 rev / 0.2986 rev/s = 3.349 seconds, but we need a 1/4 revolution for the 90 degree turn, so 3.349 / 4 = 0.83725 seconds.
Now if we apply y = y0 + v0t + 0.5*a*t^2, where a = -9.81, y = 0, v0t (vertical velocity) = 0, and plug in time, we can solve for y0, the beginning height. Rearranging yields y0 = -(0.5*-9.81*0.83725^2) = 3.438 m. Is this correct...I think this is when her center of mass is at height 0, so that would mean I need to add 1.1 m (half her body length) so that just her hands hit first, so she needs to get to a total height of 3.438+1.1 = 4.538 m...is this correct?
Thanks
Yes, your approach to solving this problem is correct. Let's go through the steps together to confirm your answer.
First, you correctly identified that the diver needs to twist 90 degrees along the trajectory. By using the conservation of angular momentum, you found the final angular velocity, wf, to be 0.2986 rev/s.
Next, you correctly determined that the time required for a quarter revolution is given by 1 rev / 0.2986 rev/s, which gives you 3.349 seconds. However, since the diver needs to turn 90 degrees and not a full 360 degrees, you divided this time by 4 to get 0.83725 seconds.
Then, you applied the kinematic equation y = y0 + v0t + 0.5*a*t^2, where a is the acceleration due to gravity (-9.81 m/s^2), y is the final height (0), v0t is the initial vertical velocity (0), and t is the time (0.83725 seconds). By plugging in these values, you correctly found y0 to be -3.438 meters.
Finally, you correctly added half of the diver's body length (1.1 meters) to account for only her hands hitting the water first. This gives a total height of 4.538 meters.
Therefore, your answer of 4.538 meters for the minimum height of the diver above the water in order to enter hands first is correct.
Great job!