A diver dives into a somersault from the platform in a fully extended position, before moving into a

Question

tuck position. Their moment of inertia at full extension is 2.54kgm2, and in a tuck position it is
0.378kgm2
. In fully extended position the somersault is initiated at 60 degrees per second rotation.
What is the angular velocity of the somersault when in the tuck position?

Answer 1

To solve this problem, we can use the principle of conservation of angular momentum.

Initial moment of inertia (I1) = 2.54 kgm^2
Final moment of inertia (I2) = 0.378 kgm^2
Initial angular velocity (ω1) = 60 degrees per second = 60 * (π/180) radians per second = π/3 radians per second

According to the principle of conservation of angular momentum:
I1 * ω1 = I2 * ω2
(2.54) * (π/3) = (0.378) * ω2
ω2 = (2.54 * π/3) / 0.378
ω2 = 6.72 radians per second

So, the angular velocity of the somersault when in the tuck position is 6.72 radians per second.