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